The Stacks Project


Tag 0DRE

96.5. Dimension theory of algebraic stacks

The main results on the dimension theory of algebraic stacks in the literature that we are aware of are those of [Osserman], which makes a study of the notions of codimension and relative dimension. We make a more detailed examination of the notion of the dimension of an algebraic stack at a point, and prove various results relating the dimension of the fibres of a morphism at a point in the source to the dimension of its source and target. We also prove a result (Lemma 96.6.4 below) which allow us (under suitable hypotheses) to compute the dimension of an algebraic stack at a point in terms of a versal ring.

While we haven't always tried to optimise our results, we have largely tried to avoid making unnecessary hypotheses. However, in some of our results, in which we compare certain properties of an algebraic stack to the properties of a versal ring to this stack at a point, we have restricted our attention to the case of algebraic stacks that are locally finitely presented over a locally Noetherian scheme base, all of whose local rings are $G$-rings. This gives us the convenience of having Artin approximation available to compare the geometry of the versal ring to the geometry of the stack itself. However, this restrictive hypothesis may not be necessary for the truth of all of the various statements that we prove. Since it is satisfied in the applications that we have in mind, though, we have been content to make it when it helps.

If $X$ is a scheme, then we define the dimension $\dim(X)$ of $X$ to be the Krull dimension of the topological space underlying $X$, while if $x$ is a point of $X$, then we define the dimension $\dim_x (X)$ of $X$ at $x$ to be the minimum of the dimensions of the open subsets $U$ of $X$ containing $x$, see Properties, Definition 27.10.1. One has the relation $\dim(X) = \sup_{x \in X} \dim_x(X)$, see Properties, Lemma 27.10.2. If $X$ is locally Noetherian, then $\dim_x(X)$ coincides with the supremum of the dimensions at $x$ of the irreducible components of $X$ passing through $x$.

If $X$ is an algebraic space and $x \in |X|$, then we define $\dim_x X = \dim_u U,$ where $U$ is any scheme admitting an étale surjection $U \to X$, and $u\in U$ is any point lying over $x$, see Properties of Spaces, Definition 57.8.1. We set $\dim(X) = \sup_{x \in |X|} \dim_x(X)$, see Properties of Spaces, Definition 57.8.2.

Remark 96.5.1. In general, the dimension of the algebraic space $X$ at a point $x$ may not coincide with the dimension of the underlying topological space $|X|$ at $x$. E.g. if $k$ is a field of characteristic zero and $X = \mathbf{A}^1_k / \mathbf{Z}$, then $X$ has dimension $1$ (the dimension of $\mathbf{A}^1_k$) at each of its points, while $|X|$ has the indiscrete topology, and hence is of Krull dimension zero. On the other hand, in Algebraic Spaces, Example 56.14.9 there is given an example of an algebraic space which is of dimension $0$ at each of its points, while $|X|$ is irreducible of Krull dimension $1$, and admits a generic point (so that the dimension of $|X|$ at any of its points is $1$); see also the discussion of this example in Properties of Spaces, Section 57.8.

On the other hand, if $X$ is a decent algebraic space, in the sense of Decent Spaces, Definition 59.6.1 (in particular, if $X$ is quasi-separated; see Decent Spaces, Section 59.6) then in fact the dimension of $X$ at $x$ does coincide with the dimension of $|X|$ at $x$; see Decent Spaces, Lemma 59.11.10.

In order to define the dimension of an algebraic stack, it will be useful to first have the notion of the relative dimension, at a point in the source, of a morphism whose source is an algebraic space, and whose target is an algebraic stack. The definition is slightly involved, just because (unlike in the case of schemes) the points of an algebraic stack, or an algebraic space, are not describable as morphisms from the spectrum of a field, but only as equivalence classes of such.

Definition 96.5.2. If $f : T \to \mathcal{X}$ is a locally of finite type morphism from an algebraic space to an algebraic stack, and if $t \in |T|$ is a point with image $x \in | \mathcal{X}|$, then we define the relative dimension of $f$ at $t$, denoted $\dim_t(T_x),$ as follows: choose a morphism $\mathop{\rm Spec} k \to \mathcal{X}$, with source the spectrum of a field, which represents $x$, and choose a point $t' \in |T \times_{\mathcal{X}} \mathop{\rm Spec} k|$ mapping to $t$ under the projection to $|T|$ (such a point $t'$ exists, by Properties of Stacks, Lemma 89.4.3); then $$ \dim_t(T_x) = \dim_{t'}(T \times_{\mathcal{X}} \mathop{\rm Spec} k ). $$

Note that since $T$ is an algebraic space and $\mathcal{X}$ is an algebraic stack, the fibre product $T \times_{\mathcal{X}} \mathop{\rm Spec} k$ is an algebraic space, and so the quantity on the right hand side of this proposed definition is in fact defined (see discussion above).

Remark 96.5.3. (1) One easily verifies (for example, by using the invariance of the relative dimension of locally of finite type morphisms of schemes under base-change; see for example Morphisms, Lemma 28.27.3 that $\dim_t(T_x)$ is well-defined, independently of the choices used to compute it.

(2) In the case that $\mathcal{X}$ is also an algebraic space, it is straightforward to confirm that this definition agrees with the definition of relative dimension given in Morphisms of Spaces, Definition 58.32.1.

We next recall the following lemma, on which our study of the dimension of a locally Noetherian algebraic stack is founded.

Lemma 96.5.4. If $f: U \to X$ is a smooth morphism of locally Noetherian algebraic spaces, and if $u \in |U|$ with image $x \in |X|$, then $$ \dim_u (U) = \dim_x(X) + \dim_{u} (U_x) $$ where $\dim_u (U_x)$ is defined via Definition 96.5.2.

Proof. See Morphisms of Spaces, Lemma 58.36.10 noting that the definition of $\dim_u (U_x)$ used here coincides with the definition used there, by Remark 96.5.3 (2). $\square$

Lemma 96.5.5. If $\mathcal{X}$ is a locally Noetherian algebraic stack and $x \in |\mathcal{X}|$. Let $U \to \mathcal{X}$ be a smooth morphism from an algebraic space to $\mathcal{X}$, let $u$ be any point of $|U|$ mapping to $x$. Then we have $$ \dim_x(\mathcal{X}) = \dim_u(U) - \dim_{u}(U_x) $$ where the relative dimension $\dim_u(U_x)$ is defined by Definition 96.5.2 and the dimension of $\mathcal{X}$ at $x$ is as in Properties of Stacks, Definition 89.12.2.

Proof. Lemma 96.5.4 can be used to verify that the right hand side $\dim_u(U) + \dim_u(U_x)$ is independent of the choice of the smooth morphism $U \to \mathcal{X}$ and $u \in |U|$. We omit the details. In particular, we may assume $U$ is a scheme. In this case we can compute $\dim_u(U_x)$ by choosing the representative of $x$ to be the composite $\mathop{\rm Spec} \kappa(u) \to U \to \mathcal{X}$, where the first morphism is the canonical one with image $u \in U$. Then, if we write $R = U \times_{\mathcal{X}} U$, and let $e : U \to R$ denote the diagonal morphism, the invariance of relative dimension under base-change shows that $\dim_u(U_x) = \dim_{e(u)}(R_u)$. Thus we see that the right hand side is equal to $\dim_u (U) - \dim_{e(u)}(R_u) = \dim_x(\mathcal{X})$ as desired. $\square$

Remark 96.5.6. For Deligne–Mumford stacks which are suitably decent (e.g. quasi-separated), it will again be the case that $\dim_x(\mathcal{X})$ coincides with the topologically defined quantity $\dim_x |\mathcal{X}|$. However, for more general Artin stacks, this will typically not be the case. For example, if $\mathcal{X} = [\mathbf{A}^1/\mathbf{G}_m]$ (over some field, with the quotient being taken with respect to the usual multiplication action of $\mathbf{G}_m$ on $\mathbf{A}^1$), then $|\mathcal{X}|$ has two points, one the specialisation of the other (corresponding to the two orbits of $\mathbf{G}_m$ on $\mathbf{A}^1$), and hence is of dimension $1$ as a topological space; but $\dim_x (\mathcal{X}) = 0$ for both points $x \in |\mathcal{X}|$. (An even more extreme example is given by the classifying space $[\mathop{\rm Spec} k/\mathbf{G}_m]$, whose dimension at its unique point is equal to $-1$.)

We can now extend Definition 96.5.2 to the context of (locally finite type) morphisms between (locally Noetherian) algebraic stacks.

Definition 96.5.7. If $f : \mathcal{T} \to \mathcal{X}$ is a locally of finite type morphism between locally Noetherian algebraic stacks, and if $t \in |\mathcal{T}|$ is a point with image $x \in |\mathcal{X}|$, then we define the relative dimension of $f$ at $t$, denoted $\dim_t(\mathcal{T}_x),$ as follows: choose a morphism $\mathop{\rm Spec} k \to \mathcal{X}$, with source the spectrum of a field, which represents $x$, and choose a point $t' \in |\mathcal{T} \times_{\mathcal{X}} \mathop{\rm Spec} k|$ mapping to $t$ under the projection to $|\mathcal{T}|$ (such a point $t'$ exists, by Properties of Stacks, Lemma 89.4.3; then $$ \dim_t(\mathcal{T}_x) = \dim_{t'}(\mathcal{T} \times_{\mathcal{X}} \mathop{\rm Spec} k ). $$

Note that since $\mathcal{T}$ is an algebraic stack and $\mathcal{X}$ is an algebraic stack, the fibre product $\mathcal{T}\times_{\mathcal{X}} \mathop{\rm Spec} k$ is an algebraic stack, which is locally Noetherian by Morphisms of Stacks, Lemma 90.17.5. Thus the quantity on the right side of this proposed definition is defined by Properties of Stacks, Definition 89.12.2.

Remark 96.5.8. Standard manipulations show that $\dim_t(\mathcal{T}_x)$ is well-defined, independently of the choices made to compute it.

We now establish some basic properties of relative dimension, which are obvious generalisations of the corresponding statements in the case of morphisms of schemes.

Lemma 96.5.9. Suppose given a Cartesian square of morphisms of locally Noetherian stacks $$ \xymatrix{ \mathcal{T}' \ar[d]\ar[r] & \mathcal{T} \ar[d] \\ \mathcal{X}' \ar[r] & \mathcal{X} } $$ in which the vertical morphisms are locally of finite type. If $t' \in |\mathcal{T}'|$, with images $t$, $x'$, and $x$ in $|\mathcal{T}|$, $|\mathcal{X}'|$, and $|\mathcal{X}|$ respectively, then $\dim_{t'}(\mathcal{T}'_{x'}) = \dim_{t}(\mathcal{T}_x).$

Proof. Both sides can (by definition) be computed as the dimension of the same fibre product. $\square$

Lemma 96.5.10. If $f: \mathcal{U} \to \mathcal{X}$ is a smooth morphism of locally Noetherian algebraic stacks, and if $u \in |\mathcal{U}|$ with image $x \in |\mathcal{X}|$, then $$ \dim_u (\mathcal{U}) = \dim_x(\mathcal{X}) + \dim_{u} (\mathcal{U}_x). $$

Proof. Choose a smooth surjective morphism $V \to \mathcal{U}$ whose source is a scheme, and let $v\in |V|$ be a point mapping to $u$. Then the composite $V \to \mathcal{U} \to \mathcal{X}$ is also smooth, and by Lemma 96.5.4 we have $\dim_x(\mathcal{X}) = \dim_v(V) - \dim_v(V_x),$ while $\dim_u(\mathcal{U}) = \dim_v(V) - \dim_v(V_u).$ Thus $$ \dim_u(\mathcal{U}) - \dim_x(\mathcal{X}) = \dim_v (V_x) - \dim_v (V_u). $$

Choose a representative $\mathop{\rm Spec} k \to \mathcal{X}$ of $x$ and choose a point $v' \in | V \times_{\mathcal{X}} \mathop{\rm Spec} k|$ lying over $v$, with image $u'$ in $|\mathcal{U}\times_{\mathcal{X}} \mathop{\rm Spec} k|$; then by definition $\dim_u(\mathcal{U}_x) = \dim_{u'}(\mathcal{U}\times_{\mathcal{X}} \mathop{\rm Spec} k),$ and $\dim_v(V_x) = \dim_{v'}(V\times_{\mathcal{X}} \mathop{\rm Spec} k).$

Now $V\times_{\mathcal{X}} \mathop{\rm Spec} k \to \mathcal{U}\times_{\mathcal{X}}\mathop{\rm Spec} k$ is a smooth surjective morphism (being the base-change of such a morphism) whose source is an algebraic space (since $V$ and $\mathop{\rm Spec} k$ are schemes, and $\mathcal{X}$ is an algebraic stack). Thus, again by definition, we have \begin{align*} \dim_{u'}(\mathcal{U}\times_{\mathcal{X}} \mathop{\rm Spec} k) & = \dim_{v'}(V\times_{\mathcal{X}} \mathop{\rm Spec} k) - \dim_{v'}(V \times_{\mathcal{X}} \mathop{\rm Spec} k)_{u'}) \\ & = \dim_v(V_x) - \dim_{v'}( (V\times_{\mathcal{X}} \mathop{\rm Spec} k)_{u'}). \end{align*} Now $V\times_{\mathcal{X}} \mathop{\rm Spec} k \cong V\times_{\mathcal{U}} (\mathcal{U}\times_{\mathcal{X}} \mathop{\rm Spec} k),$ and so Lemma 96.5.9 shows that $\dim_{v'}((V\times_{\mathcal{X}} \mathop{\rm Spec} k)_{u'}) = \dim_v(V_u).$ Putting everything together, we find that $$ \dim_u(\mathcal{U}) - \dim_x(\mathcal{X}) = \dim_u(\mathcal{U}_x), $$ as required. $\square$

Lemma 96.5.11. Let $f: \mathcal{T} \to \mathcal{X}$ be a locally of finite type morphism of algebraic stacks.

  1. The function $t \mapsto \dim_t(\mathcal{T}_{f(t)})$ is upper semi-continuous on $|\mathcal{T}|$.
  2. If $f$ is smooth, then the function $t \mapsto \dim_t(\mathcal{T}_{f(t)})$ is locally constant on $|\mathcal{T}|$.

Proof. Suppose to begin with that $\mathcal{T}$ is a scheme $T$, let $U \to \mathcal{X}$ be a smooth surjective morphism whose source is a scheme, and let $T' = T \times_{\mathcal{X}} U$. Let $f': T' \to U$ be the pull-back of $f$ over $U$, and let $g: T' \to T$ be the projection.

Lemma 96.5.9 shows that $\dim_{t'}(T'_{f'(t')}) = \dim_{g(t')}(T_{f(g(t'))}),$ for $t' \in T'$, while, since $g$ is smooth and surjective (being the base-change of a smooth surjective morphism) the map induced by $g$ on underlying topological spaces is continuous and open (by Properties of Spaces, Lemma 57.4.6), and surjective. Thus it suffices to note that part (1) for the morphism $f'$ follows from Morphisms of Spaces, Lemma 58.33.4, and part (2) from either of Morphisms, Lemma 28.28.4 or Morphisms, Lemma 28.32.12 (each of which gives the result for schemes, from which the analogous results for algebraic spaces can be deduced exactly as in Morphisms of Spaces, Lemma 58.33.4.

Now return to the general case, and choose a smooth surjective morphism $h:V \to \mathcal{T}$ whose source is a scheme. If $v \in V$, then, essentially by definition, we have $$ \dim_{h(v)}(\mathcal{T}_{f(h(v))}) = \dim_{v}(V_{f(h(v))}) - \dim_{v}(V_{h(v)}). $$ Since $V$ is a scheme, we have proved that the first of the terms on the right hand side of this equality is upper semi-continuous (and even locally constant if $f$ is smooth), while the second term is in fact locally constant. Thus their difference is upper semi-continuous (and locally constant if $f$ is smooth), and hence the function $\dim_{h(v)}(\mathcal{T}_{f(h(v))})$ is upper semi-continuous on $|V|$ (and locally constant if $f$ is smooth). Since the morphism $|V| \to |\mathcal{T}|$ is open and surjective, the lemma follows. $\square$

Before continuing with our development, we prove two lemmas related to the dimension theory of schemes.

To put the first lemma in context, we note that if $X$ is a finite-dimensional scheme, then since $\dim X$ is defined to equal the supremum of the dimensions $\dim_x X$, there exists a point $x \in X$ such that $\dim_x X = \dim X$. The following lemma shows that we may furthermore take the point $x$ to be of finite type.

Lemma 96.5.12. If $X$ is a finite-dimensional scheme, then there exists a closed (and hence finite type) point $x \in X$ such that $\dim_x X = \dim X$.

Proof. Let $d = \dim X$, and choose a maximal strictly decreasing chain of irreducible closed subsets of $X$, say \begin{equation} \tag{96.5.12.1} Z_0 \supset Z_1 \supset \ldots \supset Z_d. \end{equation} The subset $Z_d$ is a minimal irreducible closed subset of $X$, and thus any point of $Z_d$ is a generic point of $Z_d$. Since the underlying topological space of the scheme $X$ is sober, we conclude that $Z_d$ is a singleton, consisting of a single closed point $x \in X$. If $U$ is any neighbourhood of $x$, then the chain $$ U\cap Z_0 \supset U\cap Z_1 \supset \ldots \supset U\cap Z_d = Z_d = \{x\} $$ is then a strictly descending chain of irreducible closed subsets of $U$, showing that $\dim U \geq d$. Thus we find that $\dim_x X \geq d$. The other inequality being obvious, the lemma is proved. $\square$

The next lemma shows that $\dim_x X$ is a constant function on an irreducible scheme satisfying some mild additional hypotheses.

Lemma 96.5.13. If $X$ is an irreducible, Jacobson, catenary, and locally Noetherian scheme of finite dimension, then $\dim U = \dim X$ for every non-empty open subset $U$ of $X$. Equivalently, $\dim_x X$ is a constant function on $X$.

Proof. The equivalence of the two claims follows directly from the definitions. Suppose, then, that $U\subset X$ is a non-empty open subset. Certainly $\dim U \leq \dim X$, and we have to show that $\dim U \geq \dim X.$ Write $d = \dim X$, and choose a maximal strictly decreasing chain of irreducible closed subsets of $X$, say $$ X = Z_0 \supset Z_1 \supset \ldots \supset Z_d. $$ Since $X$ is Jacobson, the minimal irreducible closed subset $Z_d$ is equal to $\{x\}$ for some closed point $x$.

If $x \in U,$ then $$ U = U \cap Z_0 \supset U\cap Z_1 \supset \ldots \supset U\cap Z_d = \{x\} $$ is a strictly decreasing chain of irreducible closed subsets of $U$, and so we conclude that $\dim U \geq d$, as required. Thus we may suppose that $x \not\in U.$

Consider the flat morphism $\mathop{\rm Spec} \mathcal{O}_{X,x} \to X$. The non-empty (and hence dense) open subset $U$ of $X$ pulls back to an open subset $V \subset \mathop{\rm Spec} \mathcal{O}_{X,x}$. Replacing $U$ by a non-empty quasi-compact, and hence Noetherian, open subset, we may assume that the inclusion $U \to X$ is a quasi-compact morphism. Since the formation of scheme-theoretic images of quasi-compact morphisms commutes with flat base-change Morphisms, Lemma 28.24.15 we see that $V$ is dense in $\mathop{\rm Spec} \mathcal{O}_{X,x}$, and so in particular non-empty, and of course $x \not\in V.$ (Here we use $x$ also to denote the closed point of $\mathop{\rm Spec} \mathcal{O}_{X,x}$, since its image is equal to the given point $x \in X$.) Now $\mathop{\rm Spec} \mathcal{O}_{X,x} \setminus \{x\}$ is Jacobson Properties, Lemma 27.6.4 and hence $V$ contains a closed point $z$ of $\mathop{\rm Spec} \mathcal{O}_{X,x} \setminus \{x\}$. The closure in $X$ of the image of $z$ is then an irreducible closed subset $Z$ of $X$ containing $x$, whose intersection with $U$ is non-empty, and for which there is no irreducible closed subset properly contained in $Z$ and properly containing $\{x\}$ (because pull-back to $\mathop{\rm Spec} \mathcal{O}_{X,x}$ induces a bijection between irreducible closed subsets of $X$ containing $x$ and irreducible closed subsets of $\mathop{\rm Spec} \mathcal{O}_{X,x}$). Since $U \cap Z$ is a non-empty closed subset of $U$, it contains a point $u$ that is closed in $X$ (since $X$ is Jacobson), and since $U\cap Z$ is a non-empty (and hence dense) open subset of the irreducible set $Z$ (which contains a point not lying in $U$, namely $x$), the inclusion $\{u\} \subset U\cap Z$ is proper.

As $X$ is catenary, the chain $$ X = Z_0 \supset Z \supset \{x\} = Z_d $$ can be refined to a chain of length $d+1$, which must then be of the form $$ X = Z_0 \supset W_1 \supset \ldots \supset W_{d-1} = Z \supset \{x\} = Z_d. $$ Since $U\cap Z$ is non-empty, we then find that $$ U = U \cap Z_0 \supset U \cap W_1\supset \ldots \supset U\cap W_{d-1} = U\cap Z \supset \{u\} $$ is a strictly decreasing chain of irreducible closed subsets of $U$ of length $d+1$, showing that $\dim U \geq d$, as required. $\square$

We will prove a stack-theoretic analogue of Lemma 96.5.13 in Lemma 96.5.17 below, but before doing so, we have to introduce an additional definition, necessitated by the fact that the notion of a scheme being catenary is not an étale local one (see the example of Algebra, Remark 10.158.8 which makes it difficult to define what it means for an algebraic space or algebraic stack to be catenary (see the discussion of [Osserman, page 3]). For certain aspects of dimension theory, the following definition seems to provide a good substitute for the missing notion of a catenary algebraic stack.

Definition 96.5.14. We say that a locally Noetherian algebraic stack $\mathcal{X}$ is pseudo-catenary if there exists a smooth and surjective morphism $U \to \mathcal{X}$ whose source is a universally catenary scheme.

Example 96.5.15. If $\mathcal{X}$ is locally of finite type over a universally catenary locally Noetherian scheme $S$, and $U\to \mathcal{X}$ is a smooth surjective morphism whose source is a scheme, then the composite $U \to \mathcal{X} \to S$ is locally of finite type, and so $U$ is universally catenary Morphisms, Lemma 28.16.2. Thus $\mathcal{X}$ is pseudo-catenary.

The following lemma shows that the property of being pseudo-catenary passes through finite-type morphisms.

Lemma 96.5.16. If $\mathcal{X}$ is a pseudo-catenary locally Noetherian algebraic stack, and if $\mathcal{Y} \to \mathcal{X}$ is a locally of finite type morphism, then there exists a smooth surjective morphism $V \to \mathcal{Y}$ whose source is a universally catenary scheme; thus $\mathcal{Y}$ is again pseudo-catenary.

Proof. By assumption we may find a smooth surjective morphism $U \to \mathcal{X}$ whose source is a universally catenary scheme. The base-change $U\times_{\mathcal{X}} \mathcal{Y}$ is then an algebraic stack; let $V \to U\times_{\mathcal{X}} \mathcal{Y}$ be a smooth surjective morphism whose source is a scheme. The composite $V \to U\times_{\mathcal{X}} \mathcal{Y} \to \mathcal{Y}$ is then smooth and surjective (being a composite of smooth and surjective morphisms), while the morphism $V \to U\times_{\mathcal{X}} \mathcal{Y} \to U$ is locally of finite type (being a composite of morphisms that are locally finite type). Since $U$ is universally catenary, we see that $V$ is universally catenary (by Morphisms, Lemma 28.16.2), as claimed. $\square$

We now study the behaviour of the function $\dim_x(\mathcal{X})$ on $|\mathcal{X}|$ (for some locally Noetherian stack $\mathcal{X}$) with respect to the irreducible components of $|\mathcal{X}|$, as well as various related topics.

Lemma 96.5.17. If $\mathcal{X}$ is a Jacobson, pseudo-catenary, and locally Noetherian algebraic stack for which $|\mathcal{X}|$ is irreducible, then $\dim_x(\mathcal{X})$ is a constant function on $|\mathcal{X}|$.

Proof. It suffices to show that $\dim_x(\mathcal{X})$ is locally constant on $|\mathcal{X}|$, since it will then necessarily be constant (as $|\mathcal{X}|$ is connected, being irreducible). Since $\mathcal{X}$ is pseudo-catenary, we may find a smooth surjective morphism $U \to \mathcal{X}$ with $U$ being a univesally catenary scheme. If $\{U_i\}$ is an cover of $U$ by quasi-compact open subschemes, we may replace $U$ by $\coprod U_i,$, and it suffices to show that the function $u \mapsto \dim_{f(u)}(\mathcal{X})$ is locally constant on $U_i$. Since we check this for one $U_i$ at a time, we now drop the subscript, and write simply $U$ rather than $U_i$. Since $U$ is quasi-compact, it is the union of a finite number of irreducible components, say $T_1 \cup \ldots \cup T_n$. Note that each $T_i$ is Jacobson, catenary, and locally Noetherian, being a closed subscheme of the Jacobson, catenary, and locally Noetherian scheme $U$.

By Lemma 96.5.4, we have $\dim_{f(u)}(\mathcal{X}) = \dim_{u}(U) - \dim_{u}(U_{f(u)}).$ Lemma 96.5.11 (2) shows that the second term in the right hand expression is locally constant on $U$, as $f$ is smooth, and hence we must show that $\dim_u(U)$ is locally constant on $U$. Since $\dim_u(U)$ is the maximum of the dimensions $\dim_u T_i$, as $T_i$ ranges over the components of $U$ containing $u$, it suffices to show that if a point $u$ lies on two distinct components, say $T_i$ and $T_j$ (with $i \neq j$), then $\dim_u T_i = \dim_u T_j$, and then to note that $t\mapsto \dim_t T$ is a constant function on an irreducible Jacobson, catenary, and locally Noetherian scheme $T$ (as follows from Lemma 96.5.13).

Let $V = T_i \setminus (\bigcup_{i' \neq i} T_{i'})$ and $W = T_j \setminus (\bigcup_{i' \neq j} T_{i'})$. Then each of $V$ and $W$ is a non-empty open subset of $U$, and so each has non-empty open image in $|\mathcal{X}|$. As $|\mathcal{X}|$ is irreducible, these two non-empty open subsets of $|\mathcal{X}|$ have a non-empty intersection. Let $x$ be a point lying in this intersection, and let $v \in V$ and $w\in W$ be points mapping to $x$. We then find that $$ \dim T_i = \dim V = \dim_v (U) = \dim_x (\mathcal{X}) + \dim_v (U_x) $$ and similarly that $$ \dim T_j = \dim W = \dim_w (U) = \dim_x (\mathcal{X}) + \dim_w (U_x). $$ Since $u \mapsto \dim_u (U_{f(u)})$ is locally constant on $U$, and since $T_i \cup T_j$ is connected (being the union of two irreducible, hence connected, sets that have non-empty intersection), we see that $\dim_v (U_x) = \dim_w(U_x)$, and hence, comparing the preceding two equations, that $\dim T_i = \dim T_j$, as required. $\square$

Lemma 96.5.18. If $\mathcal{Z} \hookrightarrow \mathcal{X}$ is a closed immersion of locally Noetherian schemes, and if $z \in |\mathcal{Z}|$ has image $x \in |\mathcal{X}|$, then $\dim_z (\mathcal{Z}) \leq \dim_x(\mathcal{X})$.

Proof. Choose a smooth surjective morphism $U\to \mathcal{X}$ whose source is a scheme; the base-changed morphism $V = U\times_{\mathcal{X}} \mathcal{Z} \to \mathcal{Z}$ is then also smooth and surjective, and the projection $V \to U$ is a closed immersion. If $v \in |V|$ maps to $z \in |\mathcal{Z}|$, and if we let $u$ denote the image of $v$ in $|U|$, then clearly $\dim_v(V) \leq \dim_u(U)$, while $\dim_v (V_z) = \dim_u(U_x)$, by Lemma 96.5.9. Thus $$ \dim_z(\mathcal{Z}) = \dim_v(V) - \dim_v(V_z) \leq \dim_u(U) - \dim_u(U_x) = \dim_x(\mathcal{X}), $$ as claimed. $\square$

Lemma 96.5.19. If $\mathcal{X}$ is a locally Noetherian algebraic stack, and if $x \in |\mathcal{X}|$, then $\dim_x(\mathcal{X}) = \sup_{\mathcal{T}} \{ \dim_x(\mathcal{T}) \} $, where $\mathcal{T}$ runs over all the irreducible components of $|\mathcal{X}|$ passing through $x$ (endowed with their induced reduced structure).

Proof. Lemma 96.5.18 shows that $\dim_x (\mathcal{T}) \leq \dim_x(\mathcal{X})$ for each irreducible component $\mathcal{T}$ passing through the point $x$. Thus to prove the lemma, it suffices to show that \begin{equation} \tag{96.5.19.1} \dim_x(\mathcal{X}) \leq \sup_{\mathcal{T}} \{\dim_x(\mathcal{T})\}. \end{equation} Let $U\to\mathcal{X}$ be a smooth cover by a scheme. If $T$ is an irreducible component of $U$ then we let $\mathcal{T}$ denote the closure of its image in $\mathcal{X}$, which is an irreducible component of $\mathcal{X}$. Let $u \in U$ be a point mapping to $x$. Then we have $\dim_x(\mathcal{X})=\dim_uU-\dim_uU_x=\sup_T\dim_uT-\dim_uU_x$, where the supremum is over the irreducible components of $U$ passing through $u$. Choose a component $T$ for which the supremum is achieved, and note that $\dim_x(\mathcal{T})=\dim_uT-\dim_u T_x$. The desired inequality (96.5.19.1) now follows from the evident inequality $\dim_u T_x \leq \dim_u U_x.$ (Note that if $\mathop{\rm Spec} k \to \mathcal{X}$ is a representative of $x$, then $T\times_{\mathcal{X}} \mathop{\rm Spec} k$ is a closed subspace of $U\times_{\mathcal{X}} \mathop{\rm Spec} k$.) $\square$

Lemma 96.5.20. If $\mathcal{X}$ is a locally Noetherian algebraic stack, and if $x \in |\mathcal{X}|$, then for any open substack $\mathcal{V}$ of $\mathcal{X}$ containing $x$, there is a finite type point $x_0 \in |\mathcal{V}|$ such that $\dim_{x_0}(\mathcal{X}) = \dim_x(\mathcal{V})$.

Proof. Choose a smooth surjective morphism $f:U \to \mathcal{X}$ whose source is a scheme, and consider the function $u \mapsto \dim_{f(u)}(\mathcal{X});$ since the morphism $|U| \to |\mathcal{X}|$ induced by $f$ is open (as $f$ is smooth) as well as surjective (by assumption), and takes finite type points to finite type points (by the very definition of the finite type points of $|\mathcal{X}|$), it suffices to show that for any $u \in U$, and any open neighbourhood of $u$, there is a finite type point $u_0$ in this neighbourhood such that $\dim_{f(u_0)}(\mathcal{X}) = \dim_{f(u)}(\mathcal{X}).$ Since, with this reformulation of the problem, the surjectivity of $f$ is no longer required, we may replace $U$ by the open neighbourhood of the point $u$ in question, and thus reduce to the problem of showing that for each $u \in U$, there is a finite type point $u_0 \in U$ such that $\dim_{f(u_0)}(\mathcal{X}) = \dim_{f(u)}(\mathcal{X}).$ By Lemma 96.5.4 $\dim_{f(u)}(\mathcal{X}) = \dim_u(U) - \dim_u(U_{f(u)}),$ while $\dim_{f(u_0)}(\mathcal{X}) = \dim_{u_0}(U) - \dim_{u_0}(U_{f(u_0)}).$ Since $f$ is smooth, the expression $\dim_{u_0}(U_{f(u_0)})$ is locally constant as $u_0$ varies over $U$ (by Lemma 96.5.11 (2)), and so shrinking $U$ further around $u$ if necessary, we may assume it is constant. Thus the problem becomes to show that we may find a finite type point $u_0 \in U$ for which $\dim_{u_0}(U) = \dim_u(U)$. Since by definition $\dim_u U$ is the minimum of the dimensions $\dim V$, as $V$ ranges over the open neighbourhoods $V$ of $u$ in $U$, we may shrink $U$ down further around $u$ so that $\dim_u U = \dim U$. The existence of desired point $u_0$ then follows from Lemma 96.5.12. $\square$

Lemma 96.5.21. Let $\mathcal{T} \hookrightarrow \mathcal{X}$ be a locally of finite type monomorphism of algebraic stacks, with $\mathcal{X}$ (and thus also $\mathcal{T}$) being Jacobson, pseudo-catenary, and locally Noetherian. Suppose further that $\mathcal{T}$ is irreducible of some (finite) dimension $d$, and that $\mathcal{X}$ is reduced and of dimension less than or equal to $d$. Then there is a non-empty open substack $\mathcal{V}$ of $\mathcal{T}$ such that the induced monomorphism $\mathcal{V} \hookrightarrow \mathcal{X}$ is an open immersion which identifies $\mathcal{V}$ with an open subset of an irreducible component of $\mathcal{X}$.

Proof. Choose a smooth surjective morphism $f:U \to \mathcal{X}$ with source a scheme, necessarily reduced since $\mathcal{X}$ is, and write $U' = \mathcal{T}\times_{\mathcal{X}} U$. The base-changed morphism $U' \to U$ is a monomorphism of algebraic spaces, locally of finite type, and thus representable Morphisms of Spaces, Lemma 58.49.1 and 58.27.10; since $U$ is a scheme, so is $U'$. The projection $f': U' \to \mathcal{T}$ is again a smooth surjection. Let $u' \in U'$, with image $u \in U$. Lemma 96.5.9 shows that $\dim_{u'}(U'_{f(u')}) = \dim_u(U_{f(u)}),$ while $\dim_{f'(u')}(\mathcal{T}) =d \geq \dim_{f(u)}(\mathcal{X})$ by Lemma 96.5.17 and our assumptions on $\mathcal{T}$ and $\mathcal{X}$. Thus we see that \begin{equation} \tag{96.5.21.1} \dim_{u'} (U') = \dim_{u'} (U'_{f(u')}) + \dim_{f'(u')}(\mathcal{T}) \\ \geq \dim_u (U_{f(u)}) + \dim_{f(u)}(\mathcal{X}) = \dim_u (U). \end{equation} Since $U' \to U$ is a monomorphism, locally of finite type, it is in particular unramified, and so by the étale local structure of unramified morphisms Étale Morphisms, Lemma 40.17.3, we may find a commutative diagram $$ \xymatrix{ V' \ar[r]\ar[d] & V \ar[d] \\ U' \ar[r] & U } $$ in which the scheme $V'$ is non-empty, the vertical arrows are étale, and the upper horizontal arrow is a closed immersion. Replacing $V$ by a quasi-compact open subset whose image has non-empty intersection with the image of $U'$, and replacing $V'$ by the preimage of $V$, we may further assume that $V$ (and thus $V'$) is quasi-compact. Since $V$ is also locally Noetherian, it is thus Noetherian, and so is the union of finitely many irreducible components.

Since étale morphisms preserve pointwise dimension Descent, Lemma 34.18.2 we deduce from (96.5.21.1) that for any point $v' \in V'$, with image $v \in V$, we have $\dim_{v'}( V') \geq \dim_v(V)$. In particular, the image of $V'$ can't be contained in the intersection of two distinct irreducible components of $V$, and so we may find at least one irreducible open subset of $V$ which has non-empty intersection with $V'$; replacing $V$ by this subset, we may assume that $V$ is integral (being both reduced and irreducible). From the preceding inequality on dimensions, we conclude that the closed immersion $V' \hookrightarrow V$ is in fact an isomorphism. If we let $W$ denote the image of $V'$ in $U'$, then $W$ is a non-empty open subset of $U'$ (as étale morphisms are open), and the induced monomorphism $W \to U$ is étale (since it is so étale locally on the source, i.e. after pulling back to $V'$), and hence is an open immersion (being an étale monomorphism). Thus, if we let $\mathcal{V}$ denote the image of $W$ in $\mathcal{T}$, then $\mathcal{V}$ is a dense (equivalently, non-empty) open substack of $\mathcal{T}$, whose image is dense in an irreducible component of $\mathcal{X}$. Finally, we note that the morphism is $\mathcal{V} \to \mathcal{X}$ is smooth (since its composite with the smooth morphism $W\to \mathcal{V}$ is smooth), and also a monomorphism, and thus is an open immersion. $\square$

Lemma 96.5.22. Let $f: \mathcal{T} \to \mathcal{X}$ be a locally of finite type morphism of Jacobson, pseudo-catenary, and locally Noetherian algebraic stacks, whose source is irreducible and whose target is quasi-separated, and let $\mathcal{Z} \hookrightarrow \mathcal{X}$ denote the scheme-theoretic image of $\mathcal{T}$. Then for every finite type point $t \in |T|$, we have that $\dim_t( \mathcal{T}_{f(t)}) \geq \dim \mathcal{T} - \dim \mathcal{Z}$, and there is a non-empty (equivalently, dense) open subset of $|\mathcal{T}|$ over which equality holds.

Proof. Replacing $\mathcal{X}$ by $\mathcal{Z}$, we may and do assume that $f$ is scheme theoretically dominant, and also that $\mathcal{X}$ is irreducible. By the upper semi-continuity of fibre dimensions (Lemma 96.5.11 (1)), it suffices to prove that the equality $\dim_t( \mathcal{T}_{f(t)}) =\dim \mathcal{T} - \dim \mathcal{Z}$ holds for $t$ lying in some non-empty open substack of $\mathcal{T}$. For this reason, in the argument we are always free to replace $\mathcal{T}$ by a non-empty open substack.

Let $T' \to \mathcal{T}$ be a smooth surjective morphism whose source is a scheme, and let $T$ be a non-empty quasi-compact open subset of $T'$. Since $\mathcal{Y}$ is quasi-separated, we find that $T \to \mathcal{Y}$ is quasi-compact (by Morphisms of Stacks, Lemma 90.7.7, applied to the morphisms $T \to \mathcal{Y} \to \mathop{\rm Spec} \mathbf{Z}$). Thus, if we replace $\mathcal{T}$ by the image of $T$ in $\mathcal{T}$, then we may assume (appealing to Morphisms of Stacks, Lemma 90.7.6 that the morphism $f:\mathcal{T} \to \mathcal{X}$ is quasi-compact.

If we choose a smooth surjection $U \to \mathcal{X}$ with $U$ a scheme, then Lemma 96.3.1 ensures that we may find an irreducible open subset $V$ of $U$ such that $V \to \mathcal{X}$ is smooth and scheme-theoretically dominant. Since scheme-theoretic dominance for quasi-compact morphisms is preserved by flat base-change, the base-change $\mathcal{T} \times_{\mathcal{X}} V \to V$ of the scheme-theoretically dominant morphism $f$ is again scheme-theoretically dominant. We let $Z$ denote a scheme admitting a smooth surjection onto this fibre product; then $Z \to \mathcal{T} \times_{\mathcal{X}} V \to V$ is again scheme-theoretically dominant. Thus we may find an irreducible component $C$ of $Z$ which scheme-theoretically dominates $V$. Since the composite $Z \to \mathcal{T}\times_{\mathcal{X}} V \to \mathcal{T}$ is smooth, and since $\mathcal{T}$ is irreducible, Lemma 96.3.1 shows that any irreducible component of the source has dense image in $|\mathcal{T}|$. We now replace $C$ by a non-empty open subset $W$ which is disjoint from every other irreducible component of $Z$, and then replace $\mathcal{T}$ and $\mathcal{X}$ by the images of $W$ and $V$ (and apply Lemma 96.5.17 to see that this doesn't change the dimension of either $\mathcal{T}$ or $\mathcal{X}$). If we let $\mathcal{W}$ denote the image of the morphism $W \to \mathcal{T}\times_{\mathcal{X}} V$, then $\mathcal{W}$ is open in $\mathcal{T}\times_{\mathcal{X}} V$ (since the morphism $W \to \mathcal{T}\times_{\mathcal{X}} V$ is smooth), and is irreducible (being the image of an irreducible scheme). Thus we end up with a commutative diagram $$ \xymatrix{ W \ar[dr] \ar[r] & \mathcal{W} \ar[r] \ar[d] & V \ar[d] \\ & \mathcal{T} \ar[r] & \mathcal{X} } $$ in which $W$ and $V$ are schemes, the vertical arrows are smooth and surjective, the diagonal arrows and the left-hand upper horizontal arroware smooth, and the induced morphism $\mathcal{W} \to \mathcal{T}\times_{\mathcal{X}} V$ is an open immersion. Using this diagram, together with the definitions of the various dimensions involved in the statement of the lemma, we will reduce our verification of the lemma to the case of schemes, where it is known.

Fix $w \in |W|$ with image $w' \in |\mathcal{W}|$, image $t \in |\mathcal{T}|$, image $v$ in $|V|$, and image $x$ in $|\mathcal{X}|$. Essentially by definition (using the fact that $\mathcal{W}$ is open in $\mathcal{T}\times_{\mathcal{X}} V$, and that the fibre of a base-change is the base-change of the fibre), we obtain the equalities $$ \dim_v V_x = \dim_{w'} \mathcal{W}_t $$ and $$ \dim_t \mathcal{T}_x = \dim_{w'} \mathcal{W}_v. $$ By Lemma 96.5.4 (the diagonal arrow and right-hand vertical arrow in our diagram realise $W$ and $V$ as smooth covers by schemes of the stacks $\mathcal{T}$ and $\mathcal{X}$), we find that $$ \dim_t \mathcal{T} = \dim_w W - \dim_w W_t $$ and $$ \dim_x \mathcal{X} = \dim_v V - \dim_v V_x. $$ Combining the equalities, we find that $$ \dim_t \mathcal{T}_x - \dim_t \mathcal{T} + \dim_x \mathcal{X} = \dim_{w'} \mathcal{W}_v - \dim_w W + \dim_w W_t + \dim_v V - \dim_{w'} \mathcal{W}_t $$ Since $W \to \mathcal{W}$ is a smooth surjection, the same is true if we base-change over the morphism $\mathop{\rm Spec} \kappa(v) \to V$ (thinking of $W \to \mathcal{W}$ as a morphism over $V$), and from this smooth morphism we obtain the first of the following two equalities $$ \dim_w W_v - \dim_{w'} \mathcal{W}_v = \dim_w (W_v)_{w'} = \dim_w W_{w'}; $$ the second equality follows via a direct comparison of the two fibres involved. Similarly, if we think of $W \to \mathcal{W}$ as a morphism of schemes over $\mathcal{T}$, and base-change over some representative of the point $t \in |\mathcal{T}|$, we obtain the equalities $$ \dim_w W_t - \dim_{w'} \mathcal{W}_t = \dim_w (W_t)_{w'} = \dim_w W_{w'}. $$ Putting everything together, we find that $$ \dim_t \mathcal{T}_x - \dim_t \mathcal{T} + \dim_x \mathcal{X} = \dim_w W_v - \dim_w W + \dim_v V. $$ Our goal is to show that the left-hand side of this equality vanishes for a non-empty open subset of $t$. As $w$ varies over a non-empty open subset of $W$, its image $t \in |\mathcal{T}|$ varies over a non-empty open subset of $|\mathcal{T}|$ (as $W \to \mathcal{T}$ is smooth).

We are therefore reduced to showing that if $W\to V$ is a scheme-theoretically dominant morphism of irreducible locally Noetherian schemes that is locally of finite type, then there is a non-empty open subset of points $w\in W$ such that $\dim_w W_v =\dim_w W - \dim_v V$ (where $v$ denotes the image of $w$ in $V$). This is a standard fact, whose proof we recall for the convenience of the reader.

We may replace $W$ and $V$ by their underlying reduced subschemes without altering the validity (or not) of this equation, and thus we may assume that they are in fact integral schemes. Since $\dim_w W_v$ is locally constant on $W,$ replacing $W$ by a non-empty open subset if necessary, we may assume that $\dim_w W_v$ is constant, say equal to $d$. Choosing this open subset to be affine, we may also assume that the morphism $W\to V$ is in fact of finite type. Replacing $V$ by a non-empty open subset if necessary (and then pulling back $W$ over this open subset; the resulting pull-back is non-empty, since the flat base-change of a quasi-compact and scheme-theoretically dominant morphism remains scheme-theoretically dominant), we may furthermore assume that $W$ is flat over $V$. The morphism $W\to V$ is thus of relative dimension $d$ in the sense of Morphisms, Definition 28.28.1 and it follows from Morphisms, Lemma 28.28.6 that $\dim_w(W) = \dim_v(V) + d,$ as required. $\square$

Remark 96.5.23. We note that in the context of the preceding lemma, it need not be that $\dim \mathcal{T} \geq \dim \mathcal{Z}$; this does not contradict the inequality in the statement of the lemma, because the fibres of the morphism $f$ are again algebraic stacks, and so may have negative dimension. This is illustrated by taking $k$ to be a field, and applying the lemma to the morphism $[\mathop{\rm Spec} k/\mathbf{G}_m] \to \mathop{\rm Spec} k$.

If the morphism $f$ in the statement of the lemma is assumed to be quasi-DM (in the sense of Morphisms of Stacks, Definition 90.4.1; e.g. morphisms that are representable by algebraic spaces are quasi-DM), then the fibres of the morphism over points of the target are quasi-DM algebraic stacks, and hence are of non-negative dimension. In this case, the lemma implies that indeed $\dim \mathcal{T} \geq \dim \mathcal{Z}$. In fact, we obtain the following more general result.

Lemma 96.5.24. Let $f: \mathcal{T} \to \mathcal{X}$ be a locally of finite type morphism of Jacobson, pseudo-catenary, and locally Noetherian algebraic stacks which is quasi-DM, whose source is irreducible and whose target is quasi-separated, and let $\mathcal{Z} \hookrightarrow \mathcal{X}$ denote the scheme-theoretic image of $\mathcal{T}$. Then $\dim \mathcal{Z} \leq \dim \mathcal{T}$, and furthermore, exactly one of the following two conditions holds:

  1. for every finite type point $t \in |T|,$ we have $\dim_t(\mathcal{T}_{f(t)}) > 0,$ in which case $\dim \mathcal{Z} < \dim \mathcal{T}$; or
  2. $\mathcal{T}$ and $\mathcal{Z}$ are of the same dimension.

Proof. As was observed in the preceding remark, the dimension of a quasi-DM stack is always non-negative, from which we conclude that $\dim_t \mathcal{T}_{f(t)} \geq 0$ for all $t \in |\mathcal{T}|$, with the equality $$ \dim_t \mathcal{T}_{f(t)} = \dim_t \mathcal{T} - \dim_{f(t)} \mathcal{Z} $$ holding for a dense open subset of points $t\in |\mathcal{T}|$. $\square$

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    \section{Dimension theory of algebraic stacks}
    \label{section-dimension-of-algebraic-stacks}
    
    \noindent
    The main results on the dimension theory of algebraic stacks in the
    literature that we are aware of are those of \cite{Osserman}, which
    makes a study of the notions of codimension and relative dimension. We
    make a more detailed examination of the notion of the dimension of an
    algebraic stack at a point, and prove various results
    relating the dimension of the fibres of a morphism at a point in the source
    to the dimension of its source and target.  We also prove a result
    (Lemma \ref{lemma-dimension-formula} below) which
    allow us (under suitable hypotheses) to compute the dimension of
    an algebraic stack at a point in terms of a versal ring.
    
    \medskip\noindent
    While we haven't always tried to optimise our results, we have
    largely tried to avoid making unnecessary hypotheses.  However, in some
    of our results, in which we compare certain properties of an algebraic
    stack to the properties of a versal ring to this
    stack at a point, we have restricted our attention
    to the case of algebraic stacks that are locally finitely presented
    over a locally Noetherian scheme base, all of whose local rings are
    $G$-rings. This gives us the convenience of having Artin approximation
    available to compare the geometry of the versal ring to the geometry
    of the stack itself.  However, this restrictive hypothesis
    may not be necessary for the truth
    of all of the various statements that we prove.
    Since it is satisfied in the applications that we have in mind,
    though, we have been content to make it when it helps.
    
    \medskip\noindent
    If $X$ is a scheme, then we define the dimension $\dim(X)$ of $X$
    to be the Krull dimension of the
    topological space underlying $X$,
    while if $x$ is a point of $X$,
    then we define the dimension $\dim_x (X)$ of $X$ at $x$ to be the
    minimum of the dimensions of the open subsets $U$ of $X$ containing
    $x$, see
    Properties, Definition \ref{properties-definition-dimension}.
    One has the relation $\dim(X) = \sup_{x \in X} \dim_x(X)$, see
    Properties, Lemma \ref{properties-lemma-dimension}.
    If $X$ is locally Noetherian, then $\dim_x(X)$ coincides with the supremum
    of the dimensions at $x$
    of the irreducible components of $X$ passing through $x$.
    
    \medskip\noindent
    If $X$ is an algebraic space and $x \in |X|$,
    then we define $\dim_x X = \dim_u U,$ where $U$ is any scheme
    admitting an \'etale surjection $U \to X$,
    and $u\in U$ is any point lying over $x$, see
    Properties of Spaces, Definition
    \ref{spaces-properties-definition-dimension-at-point}.
    We set $\dim(X) = \sup_{x \in |X|} \dim_x(X)$, see
    Properties of Spaces, Definition \ref{spaces-properties-definition-dimension}.
    
    \begin{remark}
    \label{remark-dimension-algebraic-space}
    In general, the dimension of the algebraic space $X$ at a point $x$
    may not coincide with the dimension of the underlying topological space
    $|X|$ at $x$.  E.g.\ if $k$ is a field of characteristic zero and
    $X =  \mathbf{A}^1_k / \mathbf{Z}$, then $X$ has dimension $1$ (the dimension
    of $\mathbf{A}^1_k$) at each of its points,
    while $|X|$ has the indiscrete topology, and hence is of Krull
    dimension zero. On the other hand, in
    Algebraic Spaces, Example \ref{spaces-example-infinite-product}
    there is given an example of an algebraic space
    which is of dimension $0$ at each of its points, while $|X|$ is
    irreducible of Krull dimension $1$, and admits a generic point (so that the
    dimension of $|X|$ at any of its points is $1$); see also the discussion
    of this example in
    Properties of Spaces, Section \ref{spaces-properties-section-dimension}.
    
    \medskip\noindent
    On the other hand, if $X$ is a {\it decent} algebraic space, in the sense of
    Decent Spaces, Definition \ref{decent-spaces-definition-very-reasonable}
    (in particular, if $X$ is quasi-separated; see
    Decent Spaces, Section \ref{decent-spaces-section-reasonable-decent})
    then in fact the dimension of $X$ at $x$ does coincide with the dimension
    of $|X|$ at $x$; see
    Decent Spaces, Lemma \ref{decent-spaces-lemma-dimension-decent-space}.
    \end{remark}
    
    \noindent
    In order to  define the dimension of an algebraic stack,
    it will be useful to first have the notion of the relative dimension,
    at a point in the source,
    of a morphism whose source is an algebraic space,
    and whose target is an algebraic stack.  The definition is slightly
    involved, just because (unlike in the case of schemes) the points of an
    algebraic stack, or an algebraic
    space, are not describable as morphisms from the spectrum of a field,
    but only as equivalence classes of such.
    
    \begin{definition}
    \label{definition-relative-dimension}
    If $f : T \to \mathcal{X}$ is a locally of finite type morphism from an
    algebraic space to an algebraic stack,
    and if $t \in |T|$ is a point with image $x \in | \mathcal{X}|$, then we define
    {\it the relative dimension} of $f$ at $t$, denoted
    $\dim_t(T_x),$
    as follows:
    choose a morphism $\Spec k \to \mathcal{X}$, with source the spectrum of
    a field, which represents $x$, and choose a point
    $t' \in |T \times_{\mathcal{X}} \Spec k|$
    mapping to $t$ under the projection to $|T|$
    (such a point $t'$ exists, by
    Properties of Stacks, Lemma \ref{stacks-properties-lemma-points-cartesian});
    then
    $$
    \dim_t(T_x) = \dim_{t'}(T \times_{\mathcal{X}} \Spec k ).
    $$
    \end{definition}
    
    \noindent
    Note that since $T$ is an algebraic space and $\mathcal{X}$ is an
    algebraic stack, the fibre product $T \times_{\mathcal{X}} \Spec k$
    is an algebraic space, and so the quantity on the right hand side of
    this proposed definition is in fact defined (see discussion above).
    
    \begin{remark}
    \label{remark-relative-dimension}
    (1)
    One easily verifies (for example, by using the invariance
    of the relative dimension of locally of finite type morphisms of schemes
    under base-change; see for example
    Morphisms, Lemma \ref{morphisms-lemma-dimension-fibre-after-base-change}
    that $\dim_t(T_x)$ is well-defined, independently of the choices
    used to compute it.
    
    \medskip\noindent
    (2)
    In the case that $\mathcal{X}$ is also an algebraic space,
    it is straightforward to confirm that this definition agrees with
    the definition of relative dimension given in
    Morphisms of Spaces, Definition
    \ref{spaces-morphisms-definition-dimension-fibre}.
    \end{remark}
    
    \noindent
    We next recall the following lemma, on which our study of
    the dimension of a locally Noetherian algebraic stack is founded.
    
    \begin{lemma}
    \label{lemma-behaviour-of-dimensions-wrt-smooth-morphisms}
    If $f: U \to X$ is a smooth morphism of locally Noetherian algebraic
    spaces, and
    if $u \in |U|$ with image $x \in |X|$, then
    $$
    \dim_u (U) = \dim_x(X) + \dim_{u} (U_x)
    $$
    where $\dim_u (U_x)$ is defined via
    Definition \ref{definition-relative-dimension}.
    \end{lemma}
    
    \begin{proof}
    See Morphisms of Spaces, Lemma
    \ref{spaces-morphisms-lemma-smoothness-dimension-spaces}
    noting that the definition of $\dim_u (U_x)$ used here coincides with
    the definition used there, by Remark \ref{remark-relative-dimension} (2).
    \end{proof}
    
    \begin{lemma}
    \label{lemma-dimension-for-stacks}
    If $\mathcal{X}$ is a locally Noetherian algebraic stack and
    $x \in |\mathcal{X}|$. Let $U \to \mathcal{X}$ be a smooth morphism
    from an algebraic space to $\mathcal{X}$, let $u$ be any point of $|U|$
    mapping to $x$. Then we have
    $$
    \dim_x(\mathcal{X}) =  \dim_u(U) - \dim_{u}(U_x)
    $$
    where the relative dimension $\dim_u(U_x)$ is defined
    by Definition \ref{definition-relative-dimension} and the
    dimension of $\mathcal{X}$ at $x$ is as in
    Properties of Stacks, Definition
    \ref{stacks-properties-definition-dimension-at-point}.
    \end{lemma}
    
    \begin{proof}
    Lemma \ref{lemma-behaviour-of-dimensions-wrt-smooth-morphisms}
    can be used to verify that the right hand side
    $\dim_u(U) + \dim_u(U_x)$ is independent of the choice
    of the smooth morphism $U \to \mathcal{X}$ and $u \in |U|$.
    We omit the details. In particular, we may assume $U$ is a scheme.
    In this case we can compute $\dim_u(U_x)$
    by choosing the representative of $x$
    to be the composite $\Spec \kappa(u) \to U \to \mathcal{X}$, where
    the first morphism is the canonical one with image $u \in U$.
    Then, if we write $R = U \times_{\mathcal{X}} U$, and let
    $e : U \to R$ denote the diagonal morphism, the invariance of
    relative dimension under base-change shows that
    $\dim_u(U_x) = \dim_{e(u)}(R_u)$. Thus we see that
    the right hand side is equal to
    $\dim_u (U) - \dim_{e(u)}(R_u) = \dim_x(\mathcal{X})$ as desired.
    \end{proof}
    
    \begin{remark}
    \label{remark-dimension-DM}
    For Deligne--Mumford stacks which are suitably decent
    (e.g.\ quasi-separated),
    it will again be the case that $\dim_x(\mathcal{X})$ coincides with the
    topologically
    defined quantity $\dim_x |\mathcal{X}|$.  However, for more general Artin
    stacks,
    this will typically not be the case.  For example, if
    $\mathcal{X} = [\mathbf{A}^1/\mathbf{G}_m]$
    (over some field, with the quotient being taken with
    respect to the usual multiplication action of $\mathbf{G}_m$ on $\mathbf{A}^1$),
    then  $|\mathcal{X}|$ has two points, one the specialisation of the other
    (corresponding
    to the two orbits of $\mathbf{G}_m$ on $\mathbf{A}^1$), and hence is of
    dimension $1$ as
    a topological space; but $\dim_x (\mathcal{X}) = 0$ for both points
    $x \in |\mathcal{X}|$.
    (An even more extreme example is given by the classifying space
    $[\Spec k/\mathbf{G}_m]$, whose dimension at its unique point
    is equal to $-1$.)
    \end{remark}
    
    \noindent
    We can now extend Definition \ref{definition-relative-dimension}
    to the context of (locally finite type)
    morphisms between (locally Noetherian) algebraic stacks.
    
    \begin{definition}
    \label{definition-relative-dimension-for-stacks}
    If $f : \mathcal{T} \to \mathcal{X}$
    is a locally of finite type morphism between
    locally Noetherian algebraic stacks, and if
    $t \in |\mathcal{T}|$ is a point with image $x \in |\mathcal{X}|$, then
    we define the {\it relative dimension} of $f$ at $t$, denoted
    $\dim_t(\mathcal{T}_x),$ as follows:
    choose a morphism $\Spec k \to \mathcal{X}$, with source the spectrum of
    a field, which represents $x$, and choose a point
    $t' \in |\mathcal{T} \times_{\mathcal{X}} \Spec k|$
    mapping to $t$ under the projection to $|\mathcal{T}|$
    (such a point $t'$ exists, by
    Properties of Stacks, Lemma
    \ref{stacks-properties-lemma-points-cartesian}; then
    $$
    \dim_t(\mathcal{T}_x) = \dim_{t'}(\mathcal{T} \times_{\mathcal{X}} \Spec k ).
    $$
    \end{definition}
    
    \noindent
    Note that since $\mathcal{T}$ is an algebraic stack and $\mathcal{X}$ is an
    algebraic stack, the fibre product $\mathcal{T}\times_{\mathcal{X}} \Spec k$
    is an algebraic stack, which is locally Noetherian by
    Morphisms of Stacks, Lemma
    \ref{stacks-morphisms-lemma-locally-finite-type-locally-noetherian}.
    Thus the quantity on the right side of this proposed definition
    is defined by Properties of Stacks,
    Definition \ref{stacks-properties-definition-dimension-at-point}.
    
    \begin{remark}
    \label{remark-dimension-tangent-space-well-defined}
    Standard manipulations show that $\dim_t(\mathcal{T}_x)$ is well-defined,
    independently of the choices made to compute it.
    \end{remark}
    
    \noindent
    We now establish some basic properties of relative dimension, which
    are obvious generalisations of the corresponding statements in the
    case of morphisms of schemes.
    
    \begin{lemma}
    \label{lemma-base-change-invariance-of-relative-dimension}
    Suppose given
    a Cartesian square of morphisms of locally Noetherian stacks
    $$
    \xymatrix{
    \mathcal{T}' \ar[d]\ar[r] & \mathcal{T} \ar[d] \\
    \mathcal{X}' \ar[r] & \mathcal{X}
    }
    $$
    in which the vertical morphisms are locally of finite type.
    If $t' \in |\mathcal{T}'|$,
    with images $t$, $x'$, and $x$ in $|\mathcal{T}|$, $|\mathcal{X}'|$, and
    $|\mathcal{X}|$
    respectively, then $\dim_{t'}(\mathcal{T}'_{x'}) = \dim_{t}(\mathcal{T}_x).$
    \end{lemma}
    
    \begin{proof}
    Both sides can (by definition) be computed as the
    dimension of the same fibre product.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-behaviour-of-dimensions-wrt-smooth-morphisms-stacky}
    If $f: \mathcal{U} \to \mathcal{X}$ is a smooth morphism of locally Noetherian
    algebraic stacks, and
    if $u \in |\mathcal{U}|$ with image $x \in |\mathcal{X}|$,
    then
    $$
    \dim_u (\mathcal{U}) = \dim_x(\mathcal{X}) + \dim_{u} (\mathcal{U}_x).
    $$
    \end{lemma}
    
    \begin{proof}
    Choose a smooth surjective morphism $V \to \mathcal{U}$ whose source
    is a scheme, and let $v\in |V|$ be a point mapping to $u$.
    Then the composite $V \to \mathcal{U} \to \mathcal{X}$ is also smooth,
    and by Lemma \ref{lemma-behaviour-of-dimensions-wrt-smooth-morphisms}
    we have $\dim_x(\mathcal{X}) = \dim_v(V) - \dim_v(V_x),$
    while $\dim_u(\mathcal{U}) = \dim_v(V) - \dim_v(V_u).$
    Thus
    $$
    \dim_u(\mathcal{U}) - \dim_x(\mathcal{X}) = \dim_v (V_x) - \dim_v (V_u).
    $$
    
    \medskip\noindent
    Choose a representative $\Spec k \to \mathcal{X}$ of $x$
    and choose a point $v' \in | V \times_{\mathcal{X}} \Spec k|$ lying over
    $v$, with image $u'$ in $|\mathcal{U}\times_{\mathcal{X}} \Spec k|$;
    then by definition
    $\dim_u(\mathcal{U}_x) = \dim_{u'}(\mathcal{U}\times_{\mathcal{X}} \Spec k),$
    and
    $\dim_v(V_x) = \dim_{v'}(V\times_{\mathcal{X}} \Spec k).$
    
    \medskip\noindent
    Now $V\times_{\mathcal{X}} \Spec k \to \mathcal{U}\times_{\mathcal{X}}\Spec k$
    is a smooth surjective morphism (being the base-change
    of such a morphism) whose source is an algebraic space
    (since $V$ and $\Spec k$ are schemes, and $\mathcal{X}$
    is an algebraic stack).  Thus, again by definition,
    we have
    \begin{align*}
    \dim_{u'}(\mathcal{U}\times_{\mathcal{X}} \Spec k)
    & =
    \dim_{v'}(V\times_{\mathcal{X}} \Spec k) -
    \dim_{v'}(V \times_{\mathcal{X}} \Spec k)_{u'}) \\
    & = \dim_v(V_x) -
    \dim_{v'}( (V\times_{\mathcal{X}} \Spec k)_{u'}).
    \end{align*}
    Now $V\times_{\mathcal{X}} \Spec k \cong
    V\times_{\mathcal{U}} (\mathcal{U}\times_{\mathcal{X}} \Spec k),$
    and so
    Lemma \ref{lemma-base-change-invariance-of-relative-dimension}
    shows that
    $\dim_{v'}((V\times_{\mathcal{X}} \Spec k)_{u'})  = \dim_v(V_u).$
    Putting everything together, we find that
    $$
    \dim_u(\mathcal{U}) - \dim_x(\mathcal{X}) =
    \dim_u(\mathcal{U}_x),
    $$
    as required.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-relative-dimension-is-semi-continuous}
    Let $f: \mathcal{T} \to \mathcal{X}$ be a locally of finite type morphism of
    algebraic stacks.
    \begin{enumerate}
    \item
    The function $t \mapsto \dim_t(\mathcal{T}_{f(t)})$ is upper semi-continuous
    on $|\mathcal{T}|$.
    \item If $f$ is smooth, then
    the function $t \mapsto \dim_t(\mathcal{T}_{f(t)})$ is locally constant
    on $|\mathcal{T}|$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Suppose to begin with that $\mathcal{T}$ is a scheme $T$,
    let $U \to \mathcal{X}$ be a smooth surjective morphism whose source
    is a scheme, and let $T' = T \times_{\mathcal{X}} U$.
    Let $f': T' \to U$ be the pull-back of $f$ over $U$,
    and let $g: T' \to T$ be the projection.
    
    \medskip\noindent
    Lemma \ref{lemma-base-change-invariance-of-relative-dimension}
    shows that $\dim_{t'}(T'_{f'(t')}) = \dim_{g(t')}(T_{f(g(t'))}),$
    for $t' \in T'$, while,
    since $g$ is smooth and surjective (being the base-change
    of a smooth surjective morphism) the map induced by $g$ on underlying
    topological spaces is continuous and open
    (by
    Properties of Spaces, Lemma \ref{spaces-properties-lemma-topology-points}), and
    surjective. Thus it suffices to note that part (1) for the morphism $f'$
    follows from
    Morphisms of Spaces, Lemma
    \ref{spaces-morphisms-lemma-openness-bounded-dimension-fibres}, and part (2)
    from either of Morphisms, Lemma
    \ref{morphisms-lemma-flat-finite-presentation-CM-fibres-relative-dimension}
    or
    Morphisms, Lemma \ref{morphisms-lemma-smooth-omega-finite-locally-free}
    (each of which gives the result for schemes, from which
    the analogous results for algebraic spaces can
    be deduced exactly as in
    Morphisms of Spaces, Lemma
    \ref{spaces-morphisms-lemma-openness-bounded-dimension-fibres}.
    
    \medskip\noindent
    Now return to the general case,
    and choose a smooth surjective morphism
    $h:V \to \mathcal{T}$ whose source is a scheme.
    If $v \in V$, then, essentially by definition,
    we have
    $$
    \dim_{h(v)}(\mathcal{T}_{f(h(v))}) =
    \dim_{v}(V_{f(h(v))}) - \dim_{v}(V_{h(v)}).
    $$
    Since $V$ is a scheme, we have proved that the first
    of the terms on the right hand side of this equality
    is upper semi-continuous (and even locally
    constant if $f$ is smooth), while the second term is
    in fact locally constant.
    Thus their difference is upper semi-continuous
    (and locally constant if $f$ is smooth),
    and hence the function
    $\dim_{h(v)}(\mathcal{T}_{f(h(v))})$
    is upper semi-continuous on $|V|$ (and locally
    constant if $f$ is smooth).
    Since the morphism $|V| \to |\mathcal{T}|$ is open and surjective,
    the lemma follows.
    \end{proof}
    
    \noindent
    Before continuing with our development,
    we prove two lemmas related to the dimension theory of schemes.
    
    \medskip\noindent
    To put the first lemma in context,
    we note that if $X$ is a finite-dimensional scheme, then since $\dim X$
    is defined to equal the supremum of the dimensions $\dim_x X$,
    there exists a point $x \in X$ such that $\dim_x X = \dim X$.
    The following lemma shows that we may furthermore take the point
    $x$ to be of finite type.
    
    \begin{lemma}
    \label{lemma-dimension-achieved-by-finite-type-point}
    If $X$ is a finite-dimensional scheme,
    then there exists a closed (and hence finite type) point $x \in X$
    such that $\dim_x X = \dim X$.
    \end{lemma}
    
    \begin{proof}
    Let $d = \dim X$,
    and choose a maximal strictly decreasing
    chain of irreducible closed subsets of $X$,
    say
    \begin{equation}
    \label{equation-maximal-chain}
    Z_0 \supset Z_1 \supset \ldots \supset Z_d.
    \end{equation}
    The subset $Z_d$ is a minimal irreducible closed subset of $X$,
    and thus any point of $Z_d$ is a generic point of $Z_d$.
    Since the underlying topological space of the scheme $X$ is sober,
    we conclude that $Z_d$ is a singleton, consisting of a single
    closed point $x \in X$.
    If $U$ is
    any neighbourhood of $x$, then
    the chain
    $$
    U\cap Z_0 \supset U\cap Z_1 \supset \ldots \supset U\cap Z_d = Z_d =
    \{x\}
    $$
    is then a strictly descending chain of irreducible
    closed subsets of $U$, showing that $\dim U \geq d$.
    Thus we find that $\dim_x X \geq d$.  The other inequality
    being obvious, the lemma is proved.
    \end{proof}
    
    \noindent
    The next lemma shows that $\dim_x X$ is a {\it constant} function
    on an irreducible scheme satisfying some mild additional hypotheses.
    
    \begin{lemma}
    \label{lemma-constancy-of-dimension}
    If $X$ is an irreducible, Jacobson, catenary, and locally Noetherian
    scheme of finite dimension,
    then $\dim U = \dim X$ for every
    non-empty open subset $U$ of $X$.
    Equivalently, $\dim_x X$ is a constant function on $X$.
    \end{lemma}
    
    \begin{proof}
    The equivalence of the two claims follows directly from the
    definitions.   Suppose, then, that $U\subset X$ is a non-empty open
    subset.
    Certainly $\dim U \leq \dim X$, and we have to show
    that $\dim U \geq \dim X.$
    Write $d = \dim X$, and choose a maximal strictly
    decreasing chain of irreducible closed subsets
    of $X$, say
    $$
    X = Z_0 \supset Z_1 \supset \ldots \supset Z_d.
    $$
    Since $X$ is Jacobson, the minimal irreducible closed
    subset $Z_d$ is equal to $\{x\}$ for some closed
    point $x$.
    
    \medskip\noindent
    If $x \in U,$ then
    $$
    U = U \cap Z_0  \supset U\cap Z_1 \supset \ldots \supset
    U\cap Z_d = \{x\}
    $$
    is a strictly decreasing chain of irreducible closed
    subsets of $U$, and so we conclude that $\dim U \geq d$,
    as required.  Thus we may suppose that $x \not\in U.$
    
    \medskip\noindent
    Consider the flat morphism $\Spec \mathcal{O}_{X,x} \to X$.
    The non-empty (and hence dense) open subset $U$ of $X$
    pulls back to an open subset $V \subset \Spec \mathcal{O}_{X,x}$.
    Replacing $U$ by a non-empty quasi-compact, and hence
    Noetherian, open subset, we may assume that the inclusion
    $U \to X$ is a quasi-compact morphism.  Since the
    formation of scheme-theoretic images of quasi-compact
    morphisms commutes with flat base-change
    Morphisms, Lemma
    \ref{morphisms-lemma-flat-base-change-scheme-theoretic-image}
    we see that $V$ is dense in $\Spec \mathcal{O}_{X,x}$,
    and so in particular non-empty,
    and of course $x \not\in V.$  (Here we use $x$ also to denote
    the closed point of $\Spec \mathcal{O}_{X,x}$, since its image
    is equal to the given point $x \in X$.)
    Now $\Spec \mathcal{O}_{X,x} \setminus \{x\}$ is Jacobson
    Properties, Lemma
    \ref{properties-lemma-complement-closed-point-Jacobson}
    and hence $V$ contains a closed point $z$
    of $\Spec \mathcal{O}_{X,x} \setminus \{x\}$.  The closure
    in $X$ of the image of $z$ is then an irreducible
    closed subset $Z$ of $X$ containing $x$, whose intersection
    with $U$ is non-empty, and
    for which there is no irreducible closed
    subset properly contained in $Z$
    and properly containing $\{x\}$
    (because pull-back to $\Spec \mathcal{O}_{X,x}$ induces
    a bijection between irreducible closed subsets of $X$
    containing $x$ and irreducible closed subsets of $\Spec
    \mathcal{O}_{X,x}$).
    Since $U \cap Z$ is a non-empty closed subset of $U$,
    it contains a point $u$ that is closed in $X$ (since
    $X$ is Jacobson), and since $U\cap Z$
    is a non-empty (and hence dense) open subset of the irreducible set $Z$
    (which contains a point not lying in $U$, namely $x$),
    the inclusion $\{u\} \subset U\cap Z$ is proper.
    
    \medskip\noindent
    As $X$ is catenary, the chain
    $$
    X = Z_0 \supset Z \supset \{x\} = Z_d
    $$
    can be refined to a chain of length $d+1$, which must then
    be of the form
    $$
    X = Z_0 \supset W_1 \supset \ldots \supset W_{d-1} = Z \supset \{x\} = Z_d.
    $$
    Since $U\cap Z$ is non-empty, we then find that
    $$
    U = U \cap Z_0 \supset U \cap W_1\supset \ldots \supset U\cap W_{d-1}
    = U\cap Z \supset \{u\}
    $$
    is a strictly decreasing chain of irreducible closed subsets
    of $U$ of length $d+1$, showing that $\dim U \geq d$,
    as required.
    \end{proof}
    
    \noindent
    We will prove a stack-theoretic analogue
    of Lemma \ref{lemma-constancy-of-dimension}
    in Lemma \ref{lemma-irreducible-implies-equidimensional} below,
    but before doing so, we have to introduce an additional definition,
    necessitated by the fact that the notion of a scheme being catenary
    is not an \'etale local one
    (see the example of
    Algebra, Remark \ref{algebra-remark-universally-catenary-does-not-descend}
    which makes it difficult to define what it means for an algebraic
    space or algebraic stack to be catenary
    (see the discussion of \cite[page 3]{Osserman}).
    For certain aspects of dimension theory, the following
    definition seems to provide a good substitute for the missing
    notion of a catenary algebraic stack.
    
    \begin{definition}
    \label{definition-pseudo-catenary}
    We say that a locally Noetherian algebraic stack $\mathcal{X}$
    is {\it pseudo-catenary} if there exists a smooth
    and surjective morphism $U \to \mathcal{X}$ whose source is
    a universally catenary scheme.
    \end{definition}
    
    \begin{example}
    \label{example-pseudo-catenary}
    If $\mathcal{X}$ is locally of finite type over a universally
    catenary locally Noetherian scheme $S$,
    and $U\to \mathcal{X}$ is a smooth surjective morphism
    whose source is a scheme, then the composite
    $U \to \mathcal{X} \to S$ is locally of finite type,
    and so $U$ is universally catenary
    Morphisms, Lemma
    \ref{morphisms-lemma-universally-catenary-local}.
    Thus $\mathcal{X}$ is pseudo-catenary.
    \end{example}
    
    \noindent
    The following lemma shows that the property of being pseudo-catenary
    passes through finite-type morphisms.
    
    \begin{lemma}
    \label{lemma-catenary-covers}
    If $\mathcal{X}$ is a pseudo-catenary locally Noetherian algebraic
    stack, and if $\mathcal{Y} \to \mathcal{X}$ is a locally of finite type
    morphism,
    then there exists a smooth surjective morphism $V \to \mathcal{Y}$
    whose source is a universally catenary scheme; thus
    $\mathcal{Y}$ is again pseudo-catenary.
    \end{lemma}
    
    \begin{proof}
    By assumption we may find a smooth surjective morphism
    $U \to \mathcal{X}$ whose source is a universally catenary scheme.
    The base-change $U\times_{\mathcal{X}} \mathcal{Y}$ is then an algebraic
    stack; let $V \to U\times_{\mathcal{X}} \mathcal{Y}$ be a smooth
    surjective morphism whose source is a scheme.
    The composite $V \to U\times_{\mathcal{X}} \mathcal{Y} \to \mathcal{Y}$ is then
    smooth and surjective (being a composite of smooth and
    surjective morphisms), while the morphism $V \to U\times_{\mathcal{X}}
    \mathcal{Y} \to U$ is locally of finite type (being a composite
    of morphisms that are locally finite type).  Since $U$
    is universally catenary, we see that $V$ is universally catenary
    (by Morphisms, Lemma
    \ref{morphisms-lemma-universally-catenary-local}),
    as claimed.
    \end{proof}
    
    \noindent
    We now study the behaviour of the function $\dim_x(\mathcal{X})$ on
    $|\mathcal{X}|$
    (for some locally Noetherian stack $\mathcal{X}$) with respect to the
    irreducible
    components of $|\mathcal{X}|$, as well as various
    related topics.
    
    \begin{lemma}
    \label{lemma-irreducible-implies-equidimensional}
    If $\mathcal{X}$ is
    a Jacobson, pseudo-catenary, and locally Noetherian  algebraic stack
    for which $|\mathcal{X}|$ is irreducible,
    then $\dim_x(\mathcal{X})$ is a constant function on $|\mathcal{X}|$.
    \end{lemma}
    
    \begin{proof}
    It suffices to show that $\dim_x(\mathcal{X})$ is locally constant on
    $|\mathcal{X}|$,
    since it will then necessarily be constant (as $|\mathcal{X}|$ is connected,
    being irreducible).  Since $\mathcal{X}$ is pseudo-catenary,
    we may find a smooth surjective morphism $U \to \mathcal{X}$ with $U$
    being a univesally catenary scheme.  If $\{U_i\}$ is an
    cover of $U$ by quasi-compact open subschemes, we may replace
    $U$ by $\coprod U_i,$, and
    it suffices to show that
    the function $u \mapsto \dim_{f(u)}(\mathcal{X})$ is locally constant on $U_i$.
    Since we check this for one $U_i$ at a time, we now drop the subscript,
    and write simply $U$ rather than $U_i$.
    Since $U$ is quasi-compact, it
    is the union of a finite number of irreducible components,
    say $T_1 \cup \ldots \cup T_n$.  Note that each $T_i$ is Jacobson,
    catenary, and locally Noetherian,
    being a closed subscheme of the Jacobson, catenary, and locally Noetherian
    scheme $U$.
    
    \medskip\noindent
    By Lemma \ref{lemma-behaviour-of-dimensions-wrt-smooth-morphisms}, we have
    $\dim_{f(u)}(\mathcal{X}) = \dim_{u}(U) - \dim_{u}(U_{f(u)}).$
    Lemma \ref{lemma-relative-dimension-is-semi-continuous} (2)
    shows that the second term in the right hand expression is locally
    constant on $U$, as $f$ is smooth,
    and hence we must show that $\dim_u(U)$
    is locally constant on $U$.  Since $\dim_u(U)$ is the maximum
    of the dimensions $\dim_u T_i$, as $T_i$ ranges over the components
    of $U$ containing $u$, it suffices to show
    that if a point $u$ lies on two distinct components,
    say $T_i$ and $T_j$ (with $i \neq j$),
    then $\dim_u T_i = \dim_u T_j$,
    and then to note that $t\mapsto \dim_t T$ is a constant
    function on an irreducible Jacobson,
    catenary, and locally Noetherian scheme $T$
    (as follows from Lemma \ref{lemma-constancy-of-dimension}).
    
    \medskip\noindent
    Let $V = T_i \setminus (\bigcup_{i' \neq i} T_{i'})$
    and $W = T_j \setminus (\bigcup_{i' \neq j} T_{i'})$.
    Then each of $V$ and $W$ is a non-empty open subset of $U$,
    and so each has non-empty open image in $|\mathcal{X}|$.  As $|\mathcal{X}|$ is
    irreducible,
    these two non-empty open subsets of $|\mathcal{X}|$ have a non-empty
    intersection.
    Let $x$ be a point lying in this intersection, and let $v \in V$ and
    $w\in W$ be points mapping to $x$.
    We then find that
    $$
    \dim T_i = \dim V = \dim_v (U) = \dim_x (\mathcal{X}) + \dim_v (U_x)
    $$
    and similarly that
    $$
    \dim T_j = \dim W = \dim_w (U) = \dim_x (\mathcal{X}) + \dim_w (U_x).
    $$
    Since $u \mapsto \dim_u (U_{f(u)})$ is locally constant on $U$,
    and since $T_i \cup T_j$ is connected (being the union of two irreducible,
    hence connected, sets that have non-empty intersection),
    we see that $\dim_v (U_x) = \dim_w(U_x)$,
    and hence, comparing the preceding two equations,
    that $\dim T_i = \dim T_j$, as required.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-closed-immersions}
    If $\mathcal{Z} \hookrightarrow \mathcal{X}$ is a closed immersion
    of locally Noetherian schemes,
    and if $z \in |\mathcal{Z}|$ has image $x \in |\mathcal{X}|$,
    then $\dim_z (\mathcal{Z}) \leq \dim_x(\mathcal{X})$.
    \end{lemma}
    
    \begin{proof}
    Choose a smooth surjective morphism
    $U\to \mathcal{X}$ whose source is a scheme;
    the base-changed morphism
    $V = U\times_{\mathcal{X}} \mathcal{Z} \to \mathcal{Z}$
    is then also smooth and surjective, and the projection
    $V \to U$ is a closed immersion.
    If $v \in |V|$ maps to $z \in |\mathcal{Z}|$, and
    if we let $u$ denote the image of $v$ in $|U|$,
    then clearly
    $\dim_v(V) \leq \dim_u(U)$,
    while
    $\dim_v (V_z) = \dim_u(U_x)$,
    by Lemma \ref{lemma-base-change-invariance-of-relative-dimension}.
    Thus
    $$
    \dim_z(\mathcal{Z})  = \dim_v(V) - \dim_v(V_z)
    \leq \dim_u(U) - \dim_u(U_x) = \dim_x(\mathcal{X}),
    $$
    as claimed.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-dimension-via-components}
    If $\mathcal{X}$ is a locally Noetherian algebraic stack, and if
    $x \in |\mathcal{X}|$,
    then $\dim_x(\mathcal{X}) = \sup_{\mathcal{T}} \{ \dim_x(\mathcal{T}) \} $,
    where $\mathcal{T}$ runs over all the irreducible components
    of $|\mathcal{X}|$ passing through $x$ (endowed with their
    induced reduced structure).
    \end{lemma}
    
    \begin{proof}
    Lemma \ref{lemma-closed-immersions}
    shows that
    $\dim_x (\mathcal{T}) \leq \dim_x(\mathcal{X})$ for each
    irreducible component $\mathcal{T}$ passing through
    the point $x$.   Thus to prove the lemma,
    it suffices to show that
    \begin{equation}
    \label{equation-desired-inequality}
    \dim_x(\mathcal{X}) \leq
    \sup_{\mathcal{T}} \{\dim_x(\mathcal{T})\}.
    \end{equation}
    Let $U\to\mathcal{X}$ be a smooth cover by a scheme. If $T$ is an irreducible
    component of $U$ then we let $\mathcal{T}$ denote the closure of its image
    in $\mathcal{X}$, which is an irreducible component of $\mathcal{X}$. Let
    $u \in U$ be
    a point mapping to $x$. Then we have
    $\dim_x(\mathcal{X})=\dim_uU-\dim_uU_x=\sup_T\dim_uT-\dim_uU_x$, where the
    supremum is over the irreducible components of $U$ passing
    through $u$. Choose a component $T$ for which the supremum
    is achieved, and note that
    $\dim_x(\mathcal{T})=\dim_uT-\dim_u T_x$.
    The desired inequality (\ref{equation-desired-inequality})
    now follows from the evident inequality $\dim_u T_x \leq \dim_u U_x.$
    (Note that if $\Spec k \to \mathcal{X}$ is a representative of $x$,
    then $T\times_{\mathcal{X}} \Spec k$ is a closed subspace of
    $U\times_{\mathcal{X}}
    \Spec k$.)
    \end{proof}
    
    \begin{lemma}
    \label{lemma-dimension-at-finite-type-point}
    If $\mathcal{X}$ is a locally Noetherian algebraic stack, and if
    $x \in |\mathcal{X}|$, then
    for any open substack $\mathcal{V}$ of $\mathcal{X}$ containing $x$,
    there is a finite type point $x_0 \in |\mathcal{V}|$ such that
    $\dim_{x_0}(\mathcal{X}) = \dim_x(\mathcal{V})$.
    \end{lemma}
    
    \begin{proof}
    Choose a smooth surjective
    morphism  $f:U \to \mathcal{X}$ whose source is a scheme, and consider the
    function $u \mapsto \dim_{f(u)}(\mathcal{X});$
    since the morphism $|U| \to |\mathcal{X}|$ induced by $f$ is open (as $f$
    is smooth) as well as  surjective (by assumption),
    and takes finite type points to finite type points (by the very definition
    of the finite type points of $|\mathcal{X}|$),
    it suffices to show that for any $u \in U$, and any open neighbourhood of $u$,
    there is a finite type point $u_0$ in this neighbourhood such that
    $\dim_{f(u_0)}(\mathcal{X}) = \dim_{f(u)}(\mathcal{X}).$
    Since, with this reformulation
    of the problem,  the surjectivity of $f$ is no longer required,
    we may replace $U$ by the open neighbourhood of the point $u$ in question,
    and thus reduce to the problem of showing that for each $u \in U$,
    there is a finite type point $u_0 \in U$ such that
    $\dim_{f(u_0)}(\mathcal{X}) = \dim_{f(u)}(\mathcal{X}).$
    By Lemma \ref{lemma-behaviour-of-dimensions-wrt-smooth-morphisms}
    $\dim_{f(u)}(\mathcal{X}) = \dim_u(U) - \dim_u(U_{f(u)}),$
    while
    $\dim_{f(u_0)}(\mathcal{X}) = \dim_{u_0}(U) - \dim_{u_0}(U_{f(u_0)}).$
    Since $f$ is smooth, the expression $\dim_{u_0}(U_{f(u_0)})$ is locally
    constant as $u_0$ varies over $U$ (by
    Lemma \ref{lemma-relative-dimension-is-semi-continuous} (2)),
    and so shrinking $U$ further around
    $u$ if necessary, we may assume it is constant.  Thus the problem
    becomes to show that we may find a finite type point $u_0 \in U$
    for which $\dim_{u_0}(U) = \dim_u(U)$.
    Since by definition $\dim_u U$ is the minimum of the dimensions
    $\dim V$, as $V$ ranges over the open neighbourhoods $V$ of $u$
    in $U$, we may shrink $U$ down further around $u$ so that
    $\dim_u U = \dim U$.
    The existence of desired point $u_0$ then follows from
    Lemma \ref{lemma-dimension-achieved-by-finite-type-point}.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-monomorphing-a-component-in-of-the-right-dimension}
    Let $\mathcal{T} \hookrightarrow \mathcal{X}$ be a locally
    of finite type monomorphism of algebraic stacks,
    with $\mathcal{X}$ (and thus also $\mathcal{T}$)
    being Jacobson, pseudo-catenary, and locally Noetherian.
    Suppose further that $\mathcal{T}$ is irreducible
    of some (finite) dimension $d$, and that $\mathcal{X}$ is reduced
    and of dimension less
    than or equal to $d$.
    Then there is a non-empty open substack $\mathcal{V}$ of $\mathcal{T}$ such
    that the induced
    monomorphism $\mathcal{V} \hookrightarrow \mathcal{X}$ is an open immersion
    which identifies
    $\mathcal{V}$ with an open subset of an irreducible component of $\mathcal{X}$.
    \end{lemma}
    
    \begin{proof}
    Choose a smooth surjective morphism $f:U \to \mathcal{X}$ with source a scheme,
    necessarily reduced since $\mathcal{X}$ is,
    and write $U' = \mathcal{T}\times_{\mathcal{X}} U$.  The base-changed morphism
    $U' \to U$ is a monomorphism of algebraic spaces, locally of finite
    type, and thus representable
    Morphisms of Spaces, Lemma
    \ref{spaces-morphisms-lemma-locally-quasi-finite-separated-representable} and
    \ref{spaces-morphisms-lemma-monomorphism-loc-finite-type-loc-quasi-finite};
    since $U$ is a scheme, so is $U'$.
    The projection $f': U' \to \mathcal{T}$ is again a smooth surjection.
    Let $u' \in U'$, with image $u \in U$.
    Lemma \ref{lemma-base-change-invariance-of-relative-dimension}
    shows that $\dim_{u'}(U'_{f(u')}) = \dim_u(U_{f(u)}),$
    while $\dim_{f'(u')}(\mathcal{T}) =d
    \geq \dim_{f(u)}(\mathcal{X})$ by
    Lemma \ref{lemma-irreducible-implies-equidimensional}
    and our assumptions on $\mathcal{T}$ and $\mathcal{X}$.
    Thus we see that
    \begin{equation}
    \label{equation-dim-inequality}
    \dim_{u'} (U') = \dim_{u'} (U'_{f(u')}) + \dim_{f'(u')}(\mathcal{T})
    \\
    \geq \dim_u (U_{f(u)}) + \dim_{f(u)}(\mathcal{X}) = \dim_u (U).
    \end{equation}
    Since $U' \to U$ is a monomorphism, locally of finite type,
    it is in particular unramified,
    and so by the \'etale local structure of unramified morphisms
    \'Etale Morphisms, Lemma \ref{etale-lemma-finite-unramified-etale-local},
    we may find a commutative diagram
    $$
    \xymatrix{
    V' \ar[r]\ar[d] & V \ar[d] \\
    U' \ar[r] & U
    }
    $$
    in which the scheme $V'$ is non-empty,
    the vertical arrows are \'etale,
    and the upper horizontal arrow is a closed immersion.
    Replacing $V$ by a quasi-compact open subset
    whose image has non-empty intersection with the image of $U'$,
    and replacing $V'$ by the preimage of $V$, we may further
    assume that $V$ (and thus $V'$) is quasi-compact.
    Since $V$ is also locally Noetherian,
    it is thus Noetherian, and so is the union of finitely many irreducible
    components.
    
    \medskip\noindent
    Since \'etale morphisms preserve pointwise dimension
    Descent, Lemma \ref{descent-lemma-dimension-at-point-local}
    we deduce from (\ref{equation-dim-inequality})
    that for any point $v' \in V'$,
    with image $v \in V$, we have
    $\dim_{v'}( V') \geq \dim_v(V)$.
    In particular, the image of $V'$ can't be contained in the intersection
    of two distinct irreducible components of $V$, and so we may find
    at least one irreducible open subset of $V$ which has non-empty intersection
    with $V'$; replacing $V$ by this subset, we may assume that $V$ is integral
    (being both reduced and irreducible).  From the preceding inequality
    on dimensions, we conclude that the closed immersion $V' \hookrightarrow V$
    is in fact an isomorphism.
    If we let $W$ denote the image of $V'$
    in $U'$, then $W$ is a non-empty
    open subset of $U'$ (as \'etale morphisms are open),
    and the induced monomorphism $W \to U$ is \'etale
    (since it is so \'etale locally on the source, i.e.\ after pulling back
    to $V'$), and hence is an open immersion (being an \'etale monomorphism).
    Thus, if we let $\mathcal{V}$ denote the image of $W$ in $\mathcal{T}$,
    then $\mathcal{V}$ is a dense (equivalently, non-empty) open substack of
    $\mathcal{T}$,
    whose image is dense in an irreducible component of $\mathcal{X}$.
    Finally,
    we note that the morphism is $\mathcal{V} \to \mathcal{X}$ is smooth
    (since its composite
    with the smooth morphism $W\to \mathcal{V}$ is smooth),
    and also a monomorphism, and thus is an open immersion.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-dims-of-images}
    Let $f: \mathcal{T} \to \mathcal{X}$ be a locally of finite type
    morphism of Jacobson, pseudo-catenary, and locally Noetherian
    algebraic stacks,
    whose source is irreducible and whose target is quasi-separated,
    and let $\mathcal{Z} \hookrightarrow \mathcal{X}$ denote the scheme-theoretic
    image of $\mathcal{T}$.
    Then for every finite type point $t \in |T|$,
    we have that
    $\dim_t( \mathcal{T}_{f(t)}) \geq \dim \mathcal{T}  - \dim \mathcal{Z}$,
    and there is a non-empty (equivalently, dense)
    open subset of $|\mathcal{T}|$ over which equality holds.
    \end{lemma}
    
    \begin{proof}
    Replacing $\mathcal{X}$ by $\mathcal{Z}$, we may and do assume that $f$ is
    scheme theoretically dominant,
    and also that $\mathcal{X}$ is irreducible.
    By the upper semi-continuity of fibre dimensions
    (Lemma \ref{lemma-relative-dimension-is-semi-continuous} (1)),
    it suffices to prove that the equality
    $\dim_t( \mathcal{T}_{f(t)}) =\dim \mathcal{T}  - \dim \mathcal{Z}$
    holds for $t$ lying in
    some non-empty open substack of $\mathcal{T}$.
    For this reason, in the argument we are always free
    to replace $\mathcal{T}$ by a non-empty open substack.
    
    \medskip\noindent
    Let $T' \to \mathcal{T}$ be a smooth surjective morphism whose source
    is a scheme, and let $T$ be a non-empty quasi-compact open subset
    of $T'$.  Since $\mathcal{Y}$ is quasi-separated, we find
    that $T \to  \mathcal{Y}$ is quasi-compact
    (by Morphisms of Stacks, Lemma
    \ref{stacks-morphisms-lemma-quasi-compact-permanence}, applied to the morphisms
    $T \to \mathcal{Y} \to \Spec \mathbf{Z}$).
    Thus, if we replace $\mathcal{T}$ by the image of $T$ in $\mathcal{T}$,
    then we may assume (appealing to
    Morphisms of Stacks, Lemma
    \ref{stacks-morphisms-lemma-surjection-from-quasi-compact}
    that the morphism $f:\mathcal{T} \to \mathcal{X}$ is quasi-compact.
    
    \medskip\noindent
    If we choose a smooth surjection $U \to \mathcal{X}$ with $U$ a scheme,
    then Lemma \ref{lemma-map-of-components} ensures that
    we may find an irreducible open subset $V$ of $U$ such
    that $V \to \mathcal{X}$ is smooth and scheme-theoretically dominant.
    Since scheme-theoretic dominance for quasi-compact morphisms
    is preserved by flat base-change,
    the base-change $\mathcal{T} \times_{\mathcal{X}} V \to V$
    of the scheme-theoretically
    dominant morphism $f$ is again
    scheme-theoretically dominant.   We let $Z$ denote a scheme
    admitting a smooth surjection onto this fibre product;
    then $Z \to \mathcal{T} \times_{\mathcal{X}} V \to V$
    is again scheme-theoretically dominant.
    Thus we may find an irreducible
    component $C$ of $Z$ which scheme-theoretically
    dominates $V$.
    Since the composite  $Z \to \mathcal{T}\times_{\mathcal{X}} V \to \mathcal{T}$
    is smooth,
    and since $\mathcal{T}$ is irreducible,
    Lemma \ref{lemma-map-of-components} shows that any irreducible
    component of the source has dense image in $|\mathcal{T}|$.
    We now replace
    $C$ by a non-empty open subset $W$ which is disjoint from every other
    irreducible component of $Z$, and
    then replace $\mathcal{T}$ and $\mathcal{X}$ by the images of $W$
    and $V$
    (and apply Lemma \ref{lemma-irreducible-implies-equidimensional}
    to see that this
    doesn't change the dimension of either $\mathcal{T}$ or $\mathcal{X}$).
    If we let $\mathcal{W}$ denote the image of the morphism
    $W \to \mathcal{T}\times_{\mathcal{X}} V$,
    then $\mathcal{W}$ is open in $\mathcal{T}\times_{\mathcal{X}} V$ (since the
    morphism $W \to \mathcal{T}\times_{\mathcal{X}} V$ is smooth),
    and is irreducible (being the image of an irreducible
    scheme).  Thus we end up with a commutative diagram
    $$
    \xymatrix{
    W \ar[dr] \ar[r]  & \mathcal{W} \ar[r] \ar[d] & V \ar[d] \\
    & \mathcal{T} \ar[r] & \mathcal{X}
    }
    $$
    in which $W$ and $V$ are schemes,
    the vertical arrows are smooth and surjective,
    the diagonal arrows and the left-hand
    upper horizontal arroware smooth,
    and the induced morphism $\mathcal{W} \to \mathcal{T}\times_{\mathcal{X}} V$ is
    an open immersion. Using this diagram, together with the definitions
    of the various dimensions involved in
    the statement of the lemma, we will reduce our verification
    of the lemma to the case of schemes, where it is known.
    
    \medskip\noindent
    Fix $w \in |W|$ with image $w' \in |\mathcal{W}|$,
    image $t \in |\mathcal{T}|$, image $v$ in $|V|$,
    and image $x$ in $|\mathcal{X}|$.
    Essentially by definition (using the
    fact that $\mathcal{W}$ is open in $\mathcal{T}\times_{\mathcal{X}} V$, and that
    the fibre of a base-change is the base-change of the fibre),
    we obtain the equalities
    $$
    \dim_v V_x = \dim_{w'} \mathcal{W}_t
    $$
    and
    $$
    \dim_t \mathcal{T}_x = \dim_{w'} \mathcal{W}_v.
    $$
    By Lemma \ref{lemma-behaviour-of-dimensions-wrt-smooth-morphisms}
    (the diagonal arrow and right-hand vertical
    arrow in our diagram realise $W$ and $V$ as smooth covers by
    schemes of the stacks $\mathcal{T}$ and $\mathcal{X}$), we find that
    $$
    \dim_t \mathcal{T} = \dim_w W - \dim_w W_t
    $$
    and
    $$
    \dim_x \mathcal{X} = \dim_v V - \dim_v V_x.
    $$
    Combining the equalities, we find that
    $$
    \dim_t \mathcal{T}_x - \dim_t \mathcal{T} + \dim_x \mathcal{X}
    = \dim_{w'} \mathcal{W}_v - \dim_w W + \dim_w W_t + \dim_v V -
    \dim_{w'} \mathcal{W}_t
    $$
    Since $W \to \mathcal{W}$ is a smooth surjection, the same is true
    if we base-change over the morphism $\Spec \kappa(v) \to V$
    (thinking of $W \to \mathcal{W}$ as a morphism over $V$),
    and from this smooth morphism we obtain the first of the following
    two equalities
    $$
    \dim_w W_v - \dim_{w'} \mathcal{W}_v = \dim_w (W_v)_{w'} = \dim_w W_{w'};
    $$
    the second equality follows via a direct comparison of the
    two fibres involved.
    Similarly, if we think of $W \to \mathcal{W}$ as a morphism of schemes
    over $\mathcal{T}$, and base-change over some representative of the point
    $t \in |\mathcal{T}|$, we obtain the equalities
    $$
    \dim_w W_t - \dim_{w'} \mathcal{W}_t = \dim_w (W_t)_{w'} = \dim_w W_{w'}.
    $$
    Putting everything together, we find that
    $$
    \dim_t \mathcal{T}_x - \dim_t \mathcal{T} + \dim_x \mathcal{X}
    =  \dim_w W_v - \dim_w W + \dim_v V.
    $$
    Our goal is to show that the left-hand side of this equality
    vanishes for a non-empty open subset
    of $t$.  As $w$ varies over a non-empty open subset of $W$,
    its image $t \in |\mathcal{T}|$ varies over a non-empty open
    subset of $|\mathcal{T}|$ (as $W \to \mathcal{T}$ is smooth).
    
    \medskip\noindent
    We are therefore reduced to showing that if $W\to V$ is a
    scheme-theoretically dominant morphism of irreducible locally
    Noetherian schemes that is locally of finite type,
    then there is a non-empty open subset of
    points $w\in W$ such that $\dim_w W_v =\dim_w W - \dim_v V$
    (where $v$ denotes the image of $w$ in $V$).
    This is a standard fact,
    whose proof we recall for the convenience of the reader.
    
    \medskip\noindent
    We may replace $W$ and $V$ by their underlying reduced subschemes
    without altering the validity (or not) of this equation,
    and thus we may assume that they are in fact integral schemes.
    Since $\dim_w W_v$ is locally constant on $W,$ replacing $W$
    by a non-empty open subset if necessary, we may assume that $\dim_w W_v$
    is constant, say equal to $d$.  Choosing this open subset to be affine,
    we may also assume that the morphism $W\to V$ is in fact of finite type.
    Replacing $V$ by a non-empty open subset if necessary
    (and then pulling back $W$ over this open subset; the resulting pull-back
    is non-empty, since the flat base-change of a quasi-compact
    and scheme-theoretically
    dominant morphism remains scheme-theoretically dominant),
    we may furthermore assume that $W$ is flat over $V$.
    The morphism $W\to V$ is thus of relative dimension $d$
    in the sense of
    Morphisms, Definition
    \ref{morphisms-definition-relative-dimension-d}
    and it follows from
    Morphisms, Lemma \ref{morphisms-lemma-rel-dimension-dimension}
    that $\dim_w(W) = \dim_v(V) + d,$ as required.
    \end{proof}
    
    \begin{remark}
    \label{remark-negative-dimension}
    We note that in the context of the preceding lemma,
    it need not be that $\dim \mathcal{T} \geq \dim \mathcal{Z}$; this does
    not contradict the inequality in the statement of the lemma, because
    the fibres of the morphism $f$ are again algebraic stacks, and
    so may have negative dimension.  This is illustrated by taking
    $k$ to be a field, and applying the lemma to the morphism
    $[\Spec k/\mathbf{G}_m] \to \Spec k$.
    
    \medskip\noindent
    If the morphism $f$ in the statement of the lemma is assumed
    to be quasi-DM (in the sense of
    Morphisms of Stacks, Definition
    \ref{stacks-morphisms-definition-separated}; e.g.\ morphisms that are
    representable by algebraic spaces are quasi-DM),
    then the fibres of the morphism over points of the target
    are quasi-DM algebraic stacks, and hence are of non-negative
    dimension.  In this case, the lemma implies
    that indeed $\dim \mathcal{T} \geq \dim \mathcal{Z}$.  In fact, we obtain
    the following more general result.
    \end{remark}
    
    \begin{lemma}
    \label{lemma-dims-of-images-two}
    Let $f: \mathcal{T} \to \mathcal{X}$ be a locally of finite type
    morphism of Jacobson, pseudo-catenary, and locally Noetherian
    algebraic stacks
    which is quasi-DM,
    whose source is irreducible and whose target is quasi-separated,
    and let $\mathcal{Z} \hookrightarrow \mathcal{X}$ denote the scheme-theoretic
    image of $\mathcal{T}$.
    Then $\dim \mathcal{Z} \leq \dim \mathcal{T}$,
    and furthermore, exactly one of the following two conditions holds:
    \begin{enumerate}
    \item for every finite type point $t \in |T|,$
    we have
    $\dim_t(\mathcal{T}_{f(t)}) > 0,$ in which
    case $\dim \mathcal{Z} < \dim \mathcal{T}$; or
    \item   $\mathcal{T}$ and $\mathcal{Z}$
    are of the same dimension.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    As was observed in the preceding remark,
    the dimension of a quasi-DM stack is always non-negative,
    from which we conclude that $\dim_t \mathcal{T}_{f(t)} \geq 0$
    for all $t \in |\mathcal{T}|$, with the equality
    $$
    \dim_t \mathcal{T}_{f(t)} = \dim_t \mathcal{T} - \dim_{f(t)} \mathcal{Z}
    $$
    holding
    for a dense open subset of points $t\in |\mathcal{T}|$.
    \end{proof}

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