Cohomology Groups, Coefficient Groups, and Modules Two simple things that I am unable to find any good reference on by googling...
First, I wanted to be able to prove that given a space $X$, and a ring $R$, then the cohomology groups $H^{i}(X;R)$ were all necessarily $R$-modules. This should be a standard proof, but I cannot find it anywhere.
However, when looking for said such proof, I came across this line on Wikipedia:

In what follows, the coefficient group $A$ is sometimes not written. It is common to take $A$ to be a commutative ring $R$; then the cohomology groups are $R$-modules.

And that greatly confused me, as to why $H^{i} (X,R)$ need not be an $R$-module if $R$ is non-commutative.
Any references, hints, and full explanations are greatly appreciated!
 A: Okay, I think that I've figured this one out, but I could use someone to either give it their stamp of approval, or their stamp of disapproval.
Let's, for simplicity's sake, say that we are looking at simplicial cohomology. Then we have a cochain complex
$$
\dots \leftarrow C^{n+1} (X) \leftarrow C^{n} (X) \leftarrow C^{n-1} (X) \leftarrow \dots \leftarrow C^1 (X) \leftarrow C^0 (X) \leftarrow 0 .
$$
where $C^i (X) = \text{Hom}(C_i (X),R)$. Then, we can render $C^{i} (X)$ an $R$-module by defining an action $\cdot : R \times C^i (X) \rightarrow C^i (X)$ by $r \cdot \varphi=r \varphi$. Both kernels and images are submodules, this means that the induced action on $H^{i} (X,R) = \ker (C^{i} \rightarrow C^{i+1})/\text{im} (C^{i-1} \rightarrow C^{i})$ is well-defined.
The problem is that if $R$ is not commutative, then we don't necessarily have that the arrows are $R$-linear. Specifically, it is not a given that $r \cdot \delta (\varphi) = \delta (r \cdot \varphi)$.
Is that understanding correct?
