The Stacks project

Proof. See Morphisms of Spaces, Lemma 67.37.10 noting that the definition of $\dim _ u (U_ x)$ used here coincides with the definition used there, by Remark 107.5.3 (2). $\square$

Proof. Lemma 107.5.4 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 $\mathop{\mathrm{Spec}}\kappa (u) \to U \to \mathcal{X}$, 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. $\square$

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

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 107.5.4 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

Choose a representative $\mathop{\mathrm{Spec}}k \to \mathcal{X}$ of $x$ and choose a point $v' \in | V \times _{\mathcal{X}} \mathop{\mathrm{Spec}}k|$ lying over $v$, with image $u'$ in $|\mathcal{U}\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k|$; then by definition $\dim _ u(\mathcal{U}_ x) = \dim _{u'}(\mathcal{U}\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k),$ and $\dim _ v(V_ x) = \dim _{v'}(V\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k).$

Now $V\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k \to \mathcal{U}\times _{\mathcal{X}}\mathop{\mathrm{Spec}}k$ is a smooth surjective morphism (being the base-change of such a morphism) whose source is an algebraic space (since $V$ and $\mathop{\mathrm{Spec}}k$ are schemes, and $\mathcal{X}$ is an algebraic stack). Thus, again by definition, we have

Now $V\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k \cong V\times _{\mathcal{U}} (\mathcal{U}\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k),$ and so Lemma 107.5.9 shows that $\dim _{v'}((V\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k)_{u'}) = \dim _ v(V_ u).$ Putting everything together, we find that

as required. $\square$

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 107.5.9 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 66.4.6), and surjective. Thus it suffices to note that part (1) for the morphism $f'$ follows from Morphisms of Spaces, Lemma 67.34.4, and part (2) from either of Morphisms, Lemma 29.29.4 or Morphisms, Lemma 29.34.12 (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 67.34.4.

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

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. $\square$

Proof. Let $d = \dim X$, and choose a maximal strictly decreasing chain of irreducible closed subsets of $X$, say

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

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. $\square$

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

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

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 $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x} \to X$. The non-empty (and hence dense) open subset $U$ of $X$ pulls back to an open subset $V \subset \mathop{\mathrm{Spec}}\mathcal{O}_{X,x}$. 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 29.25.16 we see that $V$ is dense in $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x}$, and so in particular non-empty, and of course $x \not\in V.$ (Here we use $x$ also to denote the closed point of $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x}$, since its image is equal to the given point $x \in X$.) Now $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x} \setminus \{ x\} $ is Jacobson Properties, Lemma 28.6.4 and hence $V$ contains a closed point $z$ of $\mathop{\mathrm{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 $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x}$ induces a bijection between irreducible closed subsets of $X$ containing $x$ and irreducible closed subsets of $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x}$). 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

can be refined to a chain of length $d+1$, which must then be of the form

Since $U\cap Z$ is non-empty, we then find that

is a strictly decreasing chain of irreducible closed subsets of $U$ of length $d+1$, showing that $\dim U \geq d$, as required. $\square$

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 29.17.2), as claimed. $\square$

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 universally 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 107.5.4, we have $\dim _{f(u)}(\mathcal{X}) = \dim _{u}(U) - \dim _{u}(U_{f(u)}).$ Lemma 107.5.11 (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 107.5.13).

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

and similarly that

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. $\square$

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 107.5.9. Thus

as claimed. $\square$

Proof. Lemma 107.5.18 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

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 (107.5.19.1) now follows from the evident inequality $\dim _ u T_ x \leq \dim _ u U_ x.$ (Note that if $\mathop{\mathrm{Spec}}k \to \mathcal{X}$ is a representative of $x$, then $T\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k$ is a closed subspace of $U\times _{\mathcal{X}} \mathop{\mathrm{Spec}}k$.) $\square$

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 107.5.4 $\dim _{f(u)}(\mathcal{X}) = \dim _ u(U) - \dim _ u(U_{f(u)}),$ 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 107.5.11 (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 107.5.12. $\square$

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 67.51.1 and 67.27.10; 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 107.5.9 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 107.5.17 and our assumptions on $\mathcal{T}$ and $\mathcal{X}$. Thus we see that

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 41.17.3, we may find a commutative diagram

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 35.21.2 we deduce from (107.5.21.1) 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. $\square$

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 107.5.11 (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 101.7.7, applied to the morphisms $T \to \mathcal{Y} \to \mathop{\mathrm{Spec}}\mathbf{Z}$). Thus, if we replace $\mathcal{T}$ by the image of $T$ in $\mathcal{T}$, then we may assume (appealing to Morphisms of Stacks, Lemma 101.7.6 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 107.3.1 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 107.3.1 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 107.5.17 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

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

and

By Lemma 107.5.4 (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

and

Combining the equalities, we find that

Since $W \to \mathcal{W}$ is a smooth surjection, the same is true if we base-change over the morphism $\mathop{\mathrm{Spec}}\kappa (v) \to V$ (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

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

Putting everything together, we find that

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 29.29.1 and it follows from Morphisms, Lemma 29.29.6 that $\dim _ w(W) = \dim _ v(V) + d,$ as required. $\square$

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

holding for a dense open subset of points $t\in |\mathcal{T}|$. $\square$