The Stacks project

Lemma 101.6.2. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent:

  1. the morphism $f$ is representable by algebraic spaces,

  2. the second diagonal of $f$ is an isomorphism,

  3. the group stack $ \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is trivial over $\mathcal X$, and

  4. 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 94.15.2. 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 101.5.1 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

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!)

There are also:

  • 8 comment(s) on Section 101.6: Higher diagonals

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04YY. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 101.6.2 as pdf