Lemma 100.6.2 (04YY)—The Stacks project

Table of contents
Part 7: Algebraic Stacks
Chapter 100: Morphisms of Algebraic Stacks
Section 100.6: Higher diagonals
Lemma 100.6.2 (cite)

Lemma 100.6.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent:

the morphism $f$ is representable by algebraic spaces,
the second diagonal of $f$ is an isomorphism,
the group stack $ \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is trivial over $\mathcal X$, and
for a scheme $T$ and a morphism $x : T \to \mathcal{X}$ the kernel of $\mathit{Isom}_\mathcal {X}(x, x) \to \mathit{Isom}_\mathcal {Y}(f(x), f(x))$ is trivial.

Proof. We first prove the equivalence of (1) and (2). Namely, $f$ is representable by algebraic spaces if and only if $f$ is faithful, see Algebraic Stacks, Lemma 93.15.2. On the other hand, $f$ is faithful if and only if for every object $x$ of $\mathcal{X}$ over a scheme $T$ the functor $f$ induces an injection $\mathit{Isom}_\mathcal {X}(x, x) \to \mathit{Isom}_\mathcal {Y}(f(x), f(x))$, which happens if and only if the kernel $K$ is trivial, which happens if and only if $e : T \to K$ is an isomorphism for every $x : T \to \mathcal{X}$. Since $K = T \times _{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ as discussed above, this proves the equivalence of (1) and (2). To prove the equivalence of (2) and (3), by the discussion above, it suffices to note that a group stack is trivial if and only if its identity section is an isomorphism. Finally, the equivalence of (3) and (4) follows from the definitions: in the proof of Lemma 100.5.1 we have seen that the kernel in (4) corresponds to the fibre product $T \times _{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ over $T$. $\square$

Comments (2)

Comment #785 by Matthew Emerton on July 03, 2014 at 03:22

The following is a slightly extended comment/suggestion. Hopefully I'm not blundering.

Recall from Tag 04ZZ that a morphism of alg. stacks is a monomorphism iff its diagonal is an isomorphism of stacks.

Thus this result can be rephrased as saying that a morphism is rep'ble by alg. spaces if the diagonal is a monomorphism. In particular, it shows that condition (c) of Tag 04YQ is actually an if and only if, i.e. a morphism is rep'ble by alg. spaces iff its diagonal is a monomorphism.

It might be good to say this, since at the moment the text doesn't make any explicit connection between the current Tag and Tag 04YQ.

The hierarchy is quite nice:

monomorphism iff diagonal is an isomorphism

representable by alg. spaces iff diagonal is a monomorphism

while for an arbitrary morphism, the double diagonal is a monomorphism

so it could be fun to actually point this out explicitly, although I don't know where it would fit best; here, in the earlier section on properties of diagonals (but then the current result is in the wrong position), somewhere else, or nowhere?

(Just for fun, note also that in Tag 07Y5 (Artin's rep'bility thm.) the double diagonal is assumed to be rep'ble by alg. spaces, so this is a case where the condition that the triple diagonal be a monomorphism appears!)

Comment #803 by Johan on July 09, 2014 at 14:04

Seems right to me. I added it as a lemma in this commit. Thank you!

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