# Grothendieck topology of sets and Cech cohomology

Base concepts:

Grothendieck topology - http://en.wikipedia.org/wiki/Grothendieck_topology

Cech cohomology - https://en.wikipedia.org/wiki/%C4%8Cech_cohomology

Problem

Let $\Gamma$ be a group acting on a set $X$. I define the following (pre)topology: $\{ U_i \to U \}_{i \in I}$ is a covering if $\bigcup_{i \in I}\operatorname{Im}(U_i \to U) = U$.

Now I restrict myself to a specific covering of $X$. Let $G \subset \Gamma$ be a subgroup. Then I define $$X^G = \left\{ x \in X \mid \forall g \in G : g \cdot x = x \right\} \ .$$ Then I construct the covering $$\mathcal{U}_S = \left\{ X^{\langle g \rangle} \mid g \in \Gamma \right\} \ .$$ This is indeed a covering since $X^{\langle 1 \rangle} = X$. This covering is good, because for the Cech cohomology it produces saturated chains $$\langle g_0 \rangle \subset \langle g_0 , g_1 \rangle \subset ...$$ since $$X^{\langle g , h \rangle} = X^{\langle g \rangle} \cap X^{\langle h \rangle} \ .$$

Now let $\mathcal{A}$ be a constant sheaf sending each open set to a constant abelian $\Gamma$-group $A$.

I want to prove that the Cech cohomology (see: https://en.wikipedia.org/wiki/%C4%8Cech_cohomology ) of this sheaf can be computed from the covering $\mathcal{U}_S$ without the need to actually take a direct limit on all the possible coverings. That is, $$\check{H}^n(X,\mathcal{A}) = \check{H}^n(\mathcal{U}_S,\mathcal{A}) \ .$$

It may be the case my definition of topology on the category of $\Gamma$-set $X$ and its subsets isn't good or rigorous enough.

Motivation:

My motivation is to prove that the Cech cohomology of the constant sheaf $\mathcal{A}$ which sends every open set $U$ to the abelian $\Gamma$-group $A$ on the category of $\Gamma$-sets is bijective with the usual group cohomology, $$\check{H}^n(X,\mathcal{A}) \cong H^n(\Gamma,A) \ .$$ I already constructed a bijection between $H^n(\Gamma,A)$ and $\check{H}^n(\mathcal{U}_S,\mathcal{A})$ using saturated chains. A chain is saturated if it is of the form $$\langle g_0 \rangle \subset \langle g_0 , g_1 \rangle \subset ... \subset \langle g_0,g_1,...,g_n \rangle$$ which can be matched to the $n+1$-tuple $(g_0,g_1,...,g_n)$.

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For this to work you need $X$ to be the singleton set: consider the case where $\Gamma$ is the trivial group! –  Zhen Lin Aug 8 '13 at 12:46
If I understood your comment correctly, you say that $$\check{H}^n(X,\mathcal{A}) \not\cong \check{H}^n(\mathcal{U}_S,\mathcal{A})$$ since $X$ can have nontrivial covering (and $\mathcal{U}_S = \{ X^{\langle 1 \rangle} = X \}$). However, I still think the motivating claim about $H^n(\Gamma,A)$ is correct. –  LinAlgMan Aug 8 '13 at 13:18
My point is that you can't even get the correct answer in degree $0$: $\check{H}^0 (X, \mathcal{A})$ is always isomorphic to $H^0 (X, \mathcal{A})$, which is just the set of $\Gamma$-equivariant maps from $X$ to $A$. –  Zhen Lin Aug 8 '13 at 14:08