group action on topological space induces group action on cohomology I have two questions about group actions on topological spaces.
Suppose we have a free, properly discontinuous left group action, $G$ acts on a Hausdorff space $X$. We want to show that there is a homomorphism from $H^p(G,H^0(X,\pi^*\mathscr{F}))$ to $H^p(X/G,\mathscr{F})$, where $\mathscr{F}$ is a  sheaf on $X/G$.
Firstly, we define the $G$-action on $H^0(X,\pi^*\mathscr{F})$. From the Appendix B in Abelian varieties by Mumford, or Appendix B in Complex Abelian varieties by H.Lange, given a sheaf $\mathscr{F}$ on $X/G$, we have a left G-action on $H^0(X,\pi^*\mathscr{F})$, where $\pi:X\rightarrow X/G$. Then how does G act on $H^0(X,\pi^*\mathscr{F})$ exactly?
Is it like the following?
Suppose that $\{V_i\}_{i\in I}$ is an open covering of $X/G$ such that $\pi^{-1}(V_i)={\bigcup}_{\sigma\in G}\sigma(U_i)$, where $\pi|_{U_i}:U_i\cong V_i$ is a homeomorphism. Then $g\in G$ acts on $\prod\limits_{i\in I}\prod\limits_{\sigma\in G}\pi^*\mathscr{F}(\sigma(U_i))$ by sending $(f_{i\sigma})_{i\in I,\sigma\in G}$ to $(f_{i(g\sigma)})_{i\in I,\sigma\in G}$, since  $\pi^*\mathscr{F}(\sigma(U_i)) = \mathscr{F}(V_i)$, which will induce left $G$-action on $H^0(X,\pi^*\mathscr{F})$.
If the $G$-action is like above, then there will be a problem in H.Lange's book, P414, the first line, $$\lambda_{ij}\lambda_{jk}=\lambda_{ik}$$ We notice that $\lambda_{ij}$ is from a subset of $W_i$ to a subset of $W_j$, then right hand side from $W_i$ to $W_k$, but left hand side doesn't make any sense. If we make the equality right, it is $$\lambda_{jk}\lambda_{ij}=\lambda_{ik}$$ Then there will be a problem with the $G$-action on $H^0(X,\pi^*\mathscr{F})$ when we try to prove the theorem, which is that we cannot show the morphisms forms a morphism of complexes. In Mumford's book, P23, it says that it is easy to check. 
The second problem is how good X could be, that will allow $X/G$ to have a open covering $\{V_i\}_{i\in I}$ that has the properties in Mumford's book, P23? Mainly property 2. And how to prove it?
Thanks. And I hope I make my questions clear.
Complementary: about $\lambda_{ij}$ and $W_i$, I should change $W_i$ to $U_i$ for being consistent, then in H.Lange's book, using the language in Mumford's book:
There is an open covering $\{V_i\}_{i\in I}$ of $X/G$ such that


*

*$\pi^{-1}(V_i)={\bigcup}_{\sigma\in G}\sigma(U_i)$, where $\pi_i =\pi|_{U_i}:U_i\cong V_i$ is a homeomorphism.

*For any $i,j \in I$, there exists at most one $\lambda \in G$ such that $U_j \bigcap \lambda U_i \not = \emptyset$, call it $\lambda_{ij}$ if it exists.


We can check, if $V_i\bigcap V_j \bigcap V_k \not = \emptyset$, in the book,
$$\lambda_{ij}\lambda_{jk}=\lambda_{ik}$$
It's not correct by the directions of the maps. So we change it to $$\lambda_{jk}\lambda_{ij}=\lambda_{ik}$$
Next thing is to define a morphism of complexes. Take $\mathscr{F} = \mathscr{O}_X^\star$ for example, where $X$ is a complex manifold. For $f \in C^1(G, H^0(\mathscr{O}_X^\star))$, $\phi_1(f) \in C^1(\{V_i\}_{i\in I},\mathscr{O}_{X/G}^\star)$ defined as $\{V_i\bigcap V_j,g_{ij}\}$, $g_{ij}(x) = f(\lambda_{ij},\pi_i^{-1}(x))$. 
For $F \in C^2(G, H^0(\mathscr{O}_X^\star))$, $\phi_2(F) \in C^2(\{V_i\}_{i\in I},\mathscr{O}_{X/G}^\star)$ defined as $\{V_i\bigcap V_j \bigcap V_j,G_{ijk}\}$, $G_{ijk}(x) = F(\lambda_{ij},\lambda_{jk},\pi_i^{-1}(x))$.
Then the problem is that, it's not a moprhism of complexes, if we change the order of the $\lambda$'s equality.
 A: In general you wan the action of $G$ on $H^0(X,\pi^*\mathcal{F})$ to come from basic cohomology functoriality. Here it is : let $g:X\rightarrow X$ be any element of the group. Then there is a morphism of sheaves $\pi^*\mathcal{F}\rightarrow g_*g^*\pi^*\mathcal{F}$. We have $g^*\pi^*=\pi^*$, and taking $H^0$ : $$g^*:H^0(X,\pi^*\mathcal{F})\rightarrow H^0(X,g_*g^*\pi_*\mathcal{F})=H^0(X,g_*\pi^*\mathcal{F})=H^0(X,\pi^*\mathcal{F})$$ where the last equality is follows from the definition of the pushforward. (I denote the map on cohomology by $g^*$ because it is standard, but it may leads to some confusion as this is not the same thing as the pullback functor).
This is a right action because cohomology is contravariant : indeed the composition will come from 
$$ \pi^*\mathcal{F}\rightarrow g_*g^*\pi^*\mathcal{F}\rightarrow g_*h_*h^*g^*\pi^*\mathcal{F}=(gh)_*(gh)^*\pi^*\mathcal{F}$$
Taking $H^0$, $h^*g^*=(gh)^*$.
If you want a left action instead, you will need to look at the action of $g^{-1}$ instead. Maybe this is the thing which is troubling you as books may prefer the right action (a bit more natural) while others might prefer to stick with left ones.

Now that we have an action, we can see how it looks like, for example on open sets of the form $\bigsqcup_{\sigma\in G} \sigma(U)$ where $p:U\rightarrow V$ is an isomorphism.
If you write down all the definitions, you will see that this is what you wrote.
I don't have the books so I can't say more. I don't know what the $\lambda_{ij}$ and the $W_i$ are. Maybe you can expand your question.
