The Stacks project

99.3 Properties of diagonals

The diagonal of an algebraic stack is closely related to the $\mathit{Isom}$-sheaves, see Algebraic Stacks, Lemma 92.10.11. By the second defining property of an algebraic stack these $\mathit{Isom}$-sheaves are always algebraic spaces.

Lemma 99.3.1. Let $\mathcal{X}$ be an algebraic stack. Let $T$ be a scheme and let $x, y$ be objects of the fibre category of $\mathcal{X}$ over $T$. Then the morphism $\mathit{Isom}_\mathcal {X}(x, y) \to T$ is locally of finite type.

Proof. By Algebraic Stacks, Lemma 92.16.2 we may assume that $\mathcal{X} = [U/R]$ for some smooth groupoid in algebraic spaces. By Descent on Spaces, Lemma 72.10.9 it suffices to check the property fppf locally on $T$. Thus we may assume that $x, y$ come from morphisms $x', y' : T \to U$. By Groupoids in Spaces, Lemma 76.21.1 we see that in this case $\mathit{Isom}_\mathcal {X}(x, y) = T \times _{(y', x'), U \times _ S U} R$. Hence it suffices to prove that $R \to U \times _ S U$ is locally of finite type. This follows from the fact that the composition $s : R \to U \times _ S U \to U$ is smooth (hence locally of finite type, see Morphisms of Spaces, Lemmas 65.37.5 and 65.28.5) and Morphisms of Spaces, Lemma 65.23.6. $\square$

Lemma 99.3.2. Let $\mathcal{X}$ be an algebraic stack. Let $T$ be a scheme and let $x, y$ be objects of the fibre category of $\mathcal{X}$ over $T$. Then

  1. $\mathit{Isom}_\mathcal {X}(y, y)$ is a group algebraic space over $T$, and

  2. $\mathit{Isom}_\mathcal {X}(x, y)$ is a pseudo torsor for $\mathit{Isom}_\mathcal {X}(y, y)$ over $T$.

Proof. See Groupoids in Spaces, Definitions 76.5.1 and 76.9.1. The lemma follows immediately from the fact that $\mathcal{X}$ is a stack in groupoids. $\square$

Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The diagonal of $f$ is the morphism

\[ \Delta _ f : \mathcal{X} \longrightarrow \mathcal{X} \times _\mathcal {Y} \mathcal{X} \]

Here are two properties that every diagonal morphism has.

slogan

Lemma 99.3.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Then

  1. $\Delta _ f$ is representable by algebraic spaces, and

  2. $\Delta _ f$ is locally of finite type.

Proof. Let $T$ be a scheme and let $a : T \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ be a morphism. By definition of the fibre product and the $2$-Yoneda lemma the morphism $a$ is given by a triple $a = (x, x', \alpha )$ where $x, x'$ are objects of $\mathcal{X}$ over $T$, and $\alpha : f(x) \to f(x')$ is a morphism in the fibre category of $\mathcal{Y}$ over $T$. By definition of an algebraic stack the sheaves $\mathit{Isom}_\mathcal {X}(x, x')$ and $\mathit{Isom}_\mathcal {Y}(f(x), f(x'))$ are algebraic spaces over $T$. In this language $\alpha $ defines a section of the morphism $\mathit{Isom}_\mathcal {Y}(f(x), f(x')) \to T$. A $T'$-valued point of $\mathcal{X} \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}, a} T$ for $T' \to T$ a scheme over $T$ is the same thing as an isomorphism $x|_{T'} \to x'|_{T'}$ whose image under $f$ is $\alpha |_{T'}$. Thus we see that

99.3.3.1
\begin{equation} \label{stacks-morphisms-equation-diagonal} \vcenter { \xymatrix{ \mathcal{X} \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}, a} T \ar[d] \ar[r] & \mathit{Isom}_\mathcal {X}(x, x') \ar[d] \\ T\ar[r]^-\alpha & \mathit{Isom}_\mathcal {Y}(f(x), f(x')) } } \end{equation}

is a fibre square of sheaves over $T$. In particular we see that $\mathcal{X} \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}, a} T$ is an algebraic space which proves part (1) of the lemma.

To prove the second statement we have to show that the left vertical arrow of Diagram (99.3.3.1) is locally of finite type. By Lemma 99.3.1 the algebraic space $\mathit{Isom}_\mathcal {X}(x, x')$ and is locally of finite type over $T$. Hence the right vertical arrow of Diagram (99.3.3.1) is locally of finite type, see Morphisms of Spaces, Lemma 65.23.6. We conclude by Morphisms of Spaces, Lemma 65.23.3. $\square$

Lemma 99.3.4. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. Then

  1. $\Delta _ f$ is representable (by schemes),

  2. $\Delta _ f$ is locally of finite type,

  3. $\Delta _ f$ is a monomorphism,

  4. $\Delta _ f$ is separated, and

  5. $\Delta _ f$ is locally quasi-finite.

Proof. We have already seen in Lemma 99.3.3 that $\Delta _ f$ is representable by algebraic spaces. Hence the statements (2) – (5) make sense, see Properties of Stacks, Section 98.3. Also Lemma 99.3.3 guarantees (2) holds. Let $T \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ be a morphism and contemplate Diagram (99.3.3.1). By Algebraic Stacks, Lemma 92.9.2 the right vertical arrow is injective as a map of sheaves, i.e., a monomorphism of algebraic spaces. Hence also the morphism $T \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}} \mathcal{X} \to T$ is a monomorphism. Thus (3) holds. We already know that $T \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}} \mathcal{X} \to T$ is locally of finite type. Thus Morphisms of Spaces, Lemma 65.27.10 allows us to conclude that $T \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}} \mathcal{X} \to T$ is locally quasi-finite and separated. This proves (4) and (5). Finally, Morphisms of Spaces, Proposition 65.50.2 implies that $T \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}} \mathcal{X}$ is a scheme which proves (1). $\square$

Lemma 99.3.5. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Then the following are equivalent

  1. $f$ is separated,

  2. $\Delta _ f$ is a closed immersion,

  3. $\Delta _ f$ is proper, or

  4. $\Delta _ f$ is universally closed.

Proof. The statements “$f$ is separated”, “$\Delta _ f$ is a closed immersion”, “$\Delta _ f$ is universally closed”, and “$\Delta _ f$ is proper” refer to the notions defined in Properties of Stacks, Section 98.3. Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = \mathcal{X} \times _\mathcal {Y} V$ which is an algebraic space by assumption, and the morphism $U \to \mathcal{X}$ is surjective and smooth. By Categories, Lemma 4.31.14 and Properties of Stacks, Lemma 98.3.3 we see that for any property $P$ (as in that lemma) we have: $\Delta _ f$ has $P$ if and only if $\Delta _{U/V} : U \to U \times _ V U$ has $P$. Hence the equivalence of (2), (3) and (4) follows from Morphisms of Spaces, Lemma 65.40.9 applied to $U \to V$. Moreover, if (1) holds, then $U \to V$ is separated and we see that $\Delta _{U/V}$ is a closed immersion, i.e., (2) holds. Finally, assume (2) holds. Let $T$ be a scheme, and $a : T \to \mathcal{Y}$ a morphism. Set $T' = \mathcal{X} \times _\mathcal {Y} T$. To prove (1) we have to show that the morphism of algebraic spaces $T' \to T$ is separated. Using Categories, Lemma 4.31.14 once more we see that $\Delta _{T'/T}$ is the base change of $\Delta _ f$. Hence our assumption (2) implies that $\Delta _{T'/T}$ is a closed immersion, hence $T' \to T$ is separated as desired. $\square$

Lemma 99.3.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Then the following are equivalent

  1. $f$ is quasi-separated,

  2. $\Delta _ f$ is quasi-compact, or

  3. $\Delta _ f$ is of finite type.

Proof. The statements “$f$ is quasi-separated”, “$\Delta _ f$ is quasi-compact”, and “$\Delta _ f$ is of finite type” refer to the notions defined in Properties of Stacks, Section 98.3. Note that (2) and (3) are equivalent in view of the fact that $\Delta _ f$ is locally of finite type by Lemma 99.3.4 (and Algebraic Stacks, Lemma 92.10.9). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = \mathcal{X} \times _\mathcal {Y} V$ which is an algebraic space by assumption, and the morphism $U \to \mathcal{X}$ is surjective and smooth. By Categories, Lemma 4.31.14 and Properties of Stacks, Lemma 98.3.3 we see that we have: $\Delta _ f$ is quasi-compact if and only if $\Delta _{U/V} : U \to U \times _ V U$ is quasi-compact. If (1) holds, then $U \to V$ is quasi-separated and we see that $\Delta _{U/V}$ is quasi-compact, i.e., (2) holds. Assume (2) holds. Let $T$ be a scheme, and $a : T \to \mathcal{Y}$ a morphism. Set $T' = \mathcal{X} \times _\mathcal {Y} T$. To prove (1) we have to show that the morphism of algebraic spaces $T' \to T$ is quasi-separated. Using Categories, Lemma 4.31.14 once more we see that $\Delta _{T'/T}$ is the base change of $\Delta _ f$. Hence our assumption (2) implies that $\Delta _{T'/T}$ is quasi-compact, hence $T' \to T$ is quasi-separated as desired. $\square$

Lemma 99.3.7. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Then the following are equivalent

  1. $f$ is locally separated, and

  2. $\Delta _ f$ is an immersion.

Proof. The statements “$f$ is locally separated”, and “$\Delta _ f$ is an immersion” refer to the notions defined in Properties of Stacks, Section 98.3. Proof omitted. Hint: Argue as in the proofs of Lemmas 99.3.5 and 99.3.6. $\square$


Comments (4)

Comment #458 by Matthew Emerton on

I think in the proof of Lemma 74.3.3, there is typo in sentence 4 (the one beginning ``In this language ... ''). Namely, the target of \alpha should be Isom_{\mathcal Y}(f(x),f(x')).

Comment #461 by Toby Gee on

Glancing at the fix, it looks like there is a lower case y that should be upper case (apologies if I'm wrong - I'm viewing this on a phone).


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