I have a question about the data defining the equivariant sheaf $F$ on a scheme X from wiki: https://en.wikipedia.org/wiki/Equivariant_sheaf.
Denote by $\sigma: G \times_S X \to X$ an action of a group scheme $G$ on $X$ . Then a $O_X$-module $F$ is called equivariant if there exist in isomorphism $\phi: \sigma^* F \simeq p_2^*F$ of $\mathcal{O}_{G \times_S X}$-modules and additionally the "cocycle" condition
$$p_{23}^* \phi \circ (1_G \times \sigma)^* \phi = (m \times 1_X)^* \phi$$
is satisfied where $p_{23}, 1_G \times \sigma, m \times 1_X$ are canonical maps between $G \times G \times X$ and $G \times X$.
QUESTION: Why this data imply that on level of stalks the isomorphism $F_{gh \cdot x} \simeq F_x$ is the same as $F_{g \cdot h \cdot x} \simeq F_{h \cdot x} \simeq F_x$ (*);
namely it reflects the associativity. How to see it?
My considerations:
We have following morphisms:
- the induced isomorphism $(m \times 1_X)^* \phi$ gives the iso of $O_{G \times G \times X}$-modules
$$(m \times 1_X)^*\sigma^* F= (\sigma \circ (m \times 1_X))^*F \to (m \times 1_X)^*p_2^*F=(p_2 \circ (m \times 1_X))^*F$$
(take the contravariance $b^*a^*F= (a \circ b)^*F$ into account)
- futhermore $p_{23}^* \phi \circ (1_G \times \sigma)^* \phi$ induces conposition of isomorphisms
$$(1_G \times \sigma)^*\sigma^* F= (\sigma \circ (1_G \times \sigma))^*F \to (p_2 \circ (1_G \times \sigma))^*F=(\sigma \circ p_{23})^*F \to (p_2 \circ p_{23})^*F$$
(here we have used that $p_2 \circ (1_G \times \sigma=\sigma \circ p_{23}$)
According tot he cocycle condition the both isomorphisms 1 and 2 of $O_{G \times G \times X}$-modules coinside. By definition these isomorphisms can be given stalkwise for every given point $(g,h,x) \in G \times G \times X$. Also the general definition of pullback sheaf says that for a morphism $f:X \to Y$ and a sheaf $F$ on $Y$ we have following formula for the stalk in $z \in X$ of the pullback sheaf:
(**) $$(f^*F)_z= O_{X,z} \otimes_{O_{Y,f(z)}} F_{f(z)}$$
Now we apply this formula it to our situation for example for $f=\sigma \circ (m \times 1_X)$ and $z=(g,h,z)$ then we obtain
$$((\sigma \circ (m \times 1_X))^*F)_{(g,h,x)}= O_{G \times G \times X,(g,h,x)} \otimes_{O_{X,gh \cdot x}} F_{gh \cdot x}$$ For other $f$‘s analogously.
Now to obtain the statement (*) from we have to show that the isomorphisms 1 and 2 which by definition coinside „restrict“ in a meaningsfulway to the "right factors" of (**) such that the isos $F_{gh \cdot x} \simeq F_x$ is the same as $F_{g \cdot h \cdot x} \simeq F_{h \cdot x} \simeq F_x$ also coinside.
Could anybody help how to deduce this statement about (*) from the condition that 1 and 2 coinside?