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 $\mathscr{Y}'$.
 A: 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.
