I tried asking this at math.stackexchange but I didn't get any responses, so hopefully it's ok to try here.


I'm reading Mumford's paper "Picard Groups of Moduli Problems" and am confused about an example in the first section. I'll try to explain the situation here, but if I'm not making sense I'm talking about page 40 of the paper.


Let $\pi$ be a group, let $\mathfrak{C}$ be the category such that the objects are sets $S$ with an action of $\pi$ and morphisms $\pi$-linear maps $f:S\rightarrow T$. Maps $\{f\_\alpha: S\_\alpha\rightarrow T\}$ cover $T$ if $T=\bigcup\_{\alpha} f\_\alpha(S\_\alpha)$, so we have defined a site. Let $\langle\pi\rangle$ be $\pi$ considered as a $\pi$-set (so one forgets the group action but retains the left action of $\pi$). The claim that I'm trying to understand is that a sheaf $\mathcal{F}$ on this site is canonically determined by the $\pi$-set $M:=\mathcal{F}(\langle\pi\rangle)$. (This is a $\pi$-set by functoriality - all the elements of $\pi$ give automorphisms of $\langle\pi\rangle$ so by functoriality they give automorphisms of $M$.)


The first step of the proof is what I'm having trouble with: let $S$ be a $\pi$-set on which $\pi$ acts transitively, so we have a $\pi$-linear surjection 
$$p:\langle \pi\rangle\rightarrow S.$$
This is certainly a cover as we've defined it, so we can apply the sheaf axiom to learn... something about the corresponding map $\mathcal F(S)\rightarrow M$. Mumford claims that the sheaf axiom should imply that $\mathcal F(S)\sim M^{H}$ where $H$ is the stabilizer of some element of $S$, but I don't see why this is.



An example in Mumford's “Picard Groups of Moduli Problems”
Picking the point $s \in S$ which is the image of the identity $e \in \langle \pi \rangle$, we can identify the fiber product $\langle \pi \rangle \times\_S \langle \pi \rangle$. Any element is uniquely of the form $(g, gh)$ for some $g \in \pi$ and some element $h$ in the stabilizer $H$ of $s$. This makes the fiber product isomorphic to the disjoint union $\coprod\_{h \in H} \langle \pi \rangle$ as $\pi$-sets. Under this identification, we can understand the two projections $\coprod\_{h \in H} \langle \pi \rangle \to \langle \pi \rangle$:


* the first projection is the identity on each copy of $\langle \pi \rangle$, and
* the second projection, on the copy corresponding to $h$, is the map $\langle \pi \rangle \to \langle \pi \rangle$ given by right-multiplication by $h$.


Applying $\cal F$, the sheaf condition says that we get an equalizer diagram
$$
{\cal F}(S) \to M \rightrightarrows \prod\_{h \in H} M
$$
where the first map $M \to \prod M$ is the diagonal, and the second map $M \to \prod M$ is given by the action of $h$ on each factor. Therefore, the equalizer is precisely $M^H$.


