The Stacks project

Lemma 97.4.4. Let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. The following are equivalent

  1. $\Delta _\Delta : \mathcal{X} \to \mathcal{X} \times _{\mathcal{X} \times \mathcal{X}} \mathcal{X}$ is representable by algebraic spaces,

  2. for every $1$-morphism $\mathcal{V} \to \mathcal{X} \times \mathcal{X}$ with $\mathcal{V}$ representable (by a scheme) the fibre product $\mathcal{Y} = \mathcal{X} \times _{\Delta , \mathcal{X} \times \mathcal{X}} \mathcal{V}$ has diagonal representable by algebraic spaces.

Proof. Although this is a bit of a brain twister, it is completely formal. Namely, recall that $\mathcal{X} \times _{\mathcal{X} \times \mathcal{X}} \mathcal{X} = \mathcal{I}_\mathcal {X}$ is the inertia of $\mathcal{X}$ and that $\Delta _\Delta $ is the identity section of $\mathcal{I}_\mathcal {X}$, see Categories, Section 4.34. Thus condition (1) says the following: Given a scheme $V$, an object $x$ of $\mathcal{X}$ over $V$, and a morphism $\alpha : x \to x$ of $\mathcal{X}_ V$ the condition “$\alpha = \text{id}_ x$” defines an algebraic space over $V$. (In other words, there exists a monomorphism of algebraic spaces $W \to V$ such that a morphism of schemes $f : T \to V$ factors through $W$ if and only if $f^*\alpha = \text{id}_{f^*x}$.)

On the other hand, let $V$ be a scheme and let $x, y$ be objects of $\mathcal{X}$ over $V$. Then $(x, y)$ define a morphism $\mathcal{V} = (\mathit{Sch}/V)_{fppf} \to \mathcal{X} \times \mathcal{X}$. Next, let $h : V' \to V$ be a morphism of schemes and let $\alpha : h^*x \to h^*y$ and $\beta : h^*x \to h^*y$ be morphisms of $\mathcal{X}_{V'}$. Then $(\alpha , \beta )$ define a morphism $\mathcal{V}' = (\mathit{Sch}/V)_{fppf} \to \mathcal{Y} \times \mathcal{Y}$. Condition (2) now says that (with any choices as above) the condition “$\alpha = \beta $” defines an algebraic space over $V$.

To see the equivalence, given $(\alpha , \beta )$ as in (2) we see that (1) implies that “$\alpha ^{-1} \circ \beta = \text{id}_{h^*x}$” defines an algebraic space. The implication (2) $\Rightarrow $ (1) follows by taking $h = \text{id}_ V$ and $\beta = \text{id}_ x$. $\square$


Comments (2)

Comment #4922 by Robot0079 on

We can replace with any category fibred in groupoid, say .

Note that morphism is pull back of , hence it's both representable by algebraic spaces and diagonal representable by algebraic spaces. Therefore condition 2 is equivalent to requiring diagonal of is representable for all V.

1 implies 2 is easy, we will show the inverse below. Fix scheme T, let be a pull back of . According to previous paragraph, u becomes representable after fibred producting with any scheme V over Z. Just take and further pull back by , we see that u is representable and we win.

Comment #5190 by on

@#4922: OK, I think the proof as given is pretty perspicuous. But I did notice a typo which I fixed here.


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 07WG. Beware of the difference between the letter 'O' and the digit '0'.