Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.