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According to "Champs algébriques" by Laumon and Moret-Bailley, and $S$-groupoid is a category $\mathscr{X}$ and a functor $a: \mathscr{X} \to (\mathrm{Aff}/S)$, where $(\mathrm{Aff}/S)$ is the category of affine schemes over a fixed scheme $S$. Moreover, it satisfies the properties that make $\mathscr{X}$ a fibered category over $(\mathrm{Aff}/S)$ where each fiber category $\mathscr{X}_U$ is a groupoid.

Then the authors assert that a morphism $F: \mathscr{X} \to \mathscr{Y}$ is a monomorphism if each restriction to the fiber categories $F_U: \mathscr{X}_U \to \mathscr{Y}_U$ is fully faithful. My question is the following: why is it not sufficient that each $F_U$ be only faithful? Why is fullness necessary?

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You wrote "if," not "if and only if," so are you sure the authors are asserting that fullness is necessary? –  Qiaochu Yuan Jun 5 '12 at 18:04
    
I believe they mean "if and only if", because that it how it is used in later comments. However, it is possible that this is just their definition of a monomorphism -- in this case, I'd like to see how to reconcile this definition with the standard one. –  Michael Kasa Jun 5 '12 at 18:13

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The appropriate notion of "monomorphism" between groupoids is a fully faithful imbedding because the existence of isomorphisms between objects is a type of quotienting (which is more subtle then declaring two objects are equal). If you have a map of groupoids which is only faithful, then there might be more quotienting in the target which was not done in the source.

For instance, there is a map from the stack $\ast$ (which assigns any scheme the point) to the stack $BG$ (which assigns principal $G$-bundles) given by the trivial $G$-bundle. Then this map is a faithful functor $\ast \to BG$, but it is very far from being a monomorphism because there are lots of automorphisms in $BG$ which don't come from $\ast$. In fact, $BG$ is the "stacky" quotient of $\ast$ modulo the action of $G$.

A bit more motivation: given a map of groupoids (or $S$-groupoids) $C \to D$, then $C \to D$ is fully faithful if and only if $C \to C \times_D C$ (that's the homotopy or 2-fibered product) is an equivalence. This coincides with the "classical" remark that a map $X \to Y$ is a category if and only if $X \to X \times_Y X$ is an isomorphism, but in 2-category land.

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