# Descending a monomorphism of stacks

The question is about Proposition 3.8.1 in Laumon and Moret-Bailly book on algebraic stacks.

Let $S$ be a scheme and let $F: \mathscr{X} \rightarrow \mathscr{Y}$ be a morphism of $S$-stacks (for the etale topology, say). Let $F': \mathscr{X}' \rightarrow \mathscr{Y}'$ be the base change of $F$ along an epimorphism of $S$-stacks $Q\colon \mathscr{Y}' \rightarrow \mathscr{Y}$. Suppose that $F'$ is a monomorphism. Is $F$ then necessarily a monomorphism, too?

This is claimed in the reference cited above, but the proof given there is unclear to me. In the proof they seem to use that $\mathscr{X}' \times_{\mathscr{Y}'} \mathscr{X}' \rightarrow \mathscr{X} \times_{\mathscr{Y}} \mathscr{X}$ is an epimorphism, but I cannot see why this is so, the subtlety being that these are $2$-fiber products, so objects take into account isomorphisms in $\mathscr{Y}'$ and $\mathscr{Y}$ and we cannot a priori lift an isomorphism in a fiber category of $\mathscr{Y}$ to $\mathscr{Y}'$.

The proof given in Laumon and Moret-Bailly is clear and does not seem to use your claim.

Denote by $G=\Delta_F :\mathcal X \to \mathcal X \times_{\mathcal Y} \mathcal X$ the diagonal morphism. The main points used are

A. $F$ is representable iff $\Delta_F$ is a mono (not really used, but good to know)

B. $F$ is a mono iff $\Delta_F$ is an iso

C. $\Delta_G$ is always a mono

for proofs of these facts you can have a look at http://stacks.math.columbia.edu/tag/0AHJ

D. The formation of $\Delta_F$ commutes with based change

E. A morphism of stacks is an iso iff it is a mono and an epi

F. $F'$ is an epi iff $F$ is

And this is enough :

First case : assume $F$ is representable, i.e. $\Delta_F$ is a mono. Then if $F'$ is a mono, $\Delta_{F'}$ is an iso, hence an epi, hence $\Delta_F$ is a epi as well, hence an iso, so $F$ is a mono.

General case : First note that since $F'$ is a mono by hypothesis, $G'$ is an iso. Since $G$ is representable, we deduce from the first case that $G$ is an iso as well, that is, $F$ is a mono.

• Thank you, but I don't completely understand your argument for the same reason that made me question the original proof. Namely, your F. is not an 'if and only if' in general (to get $F'$ epi $\Rightarrow$ $F$ epi you need that $Q$ is an epi). This then seems to be a problem at the step "$\Delta_{F'}$ is epi, hence $\Delta_F$ is epi" precisely for the reason that we don't know that the new $Q$ (written out in my question) is an epi! Could you clarify? – O-Ren Ishii Apr 25 '15 at 16:07
• You are simply confused, because there is no such new $Q$. That is my point D. – Niels Apr 25 '15 at 19:49
• You are right! The key point that I overlooked is that $\mathscr{X}' \times_{\mathscr{Y}'} \mathscr{X}' \rightarrow \mathscr{X} \times_{\mathscr{Y}} \mathscr{X}$ is itself a base change along $\mathscr{Y}' \rightarrow \mathscr{Y}$. Thank you very much for taking time to help me! This has been extremely useful. (I would vote your answer up if I had enough reputation to do that.) – O-Ren Ishii Apr 25 '15 at 20:29