Equivariant cohomology vs. invariant cohomology vs. cohomology of quotient space

Question

Given a space $X$ and an action of a group $G$ on $X$, the $G$-invariant cochains with coefficients in an Abelian group $A$ define a sub-cocomplex $\mathcal{C}^{\bullet}_G$ of the cocomplex $\mathcal{C}^{\bullet}$ of cochains with coefficients in $A$. I will call the cohomology $H^{\bullet}(\mathcal{C}_G)$ the "invariant cohomology" of $X$. What is the relationship between the follwing three objects: (i) the invariant cohomology, (ii) the usual equivariant cohomology $\mathcal{H}_G^{\bullet}(X, A)$, and (iii) the cohomology of the quotient space $H^{\bullet}(X/G,A)$? Note that I am most interested in the case where $A$ is a finite Abelian group.

I think that there are always homomorphisms $$H^n(X/G,A) \to H^n(\mathcal{C}_G) \to \mathcal{H}_G^n(X,A).$$ The first map comes from pulling back the projection $X \to X/G$, and the second can be seen from the interpretation of equivariant cohomology as the total cohomology of the double cocomplex of group cochains of $G$ valued in $\mathcal{C}^{\bullet}$. But how does one characterize the kernel and image of these maps? Is there a simple statement if $X$ is contractible, say?

[Edit: regarding the relationship between $H^n(X/G,A)$ and $H^n(\mathcal{C}_G)$, please see the comments on Mark Grant's answer. The upshot is that with simplicial cochains, $H^n(\mathcal{C}_G)$ seems to depend on the choice of triangulation. But, if one defines $\mathcal{C}_G$ in a suitably triangulation-independent way, i.e. either singular cochains, or simplicial cochains on the abstract simplicial complex containing all simplex embeddings into X, then $H^n(\mathcal{C}_G)$ and $H^n(X/G,A)$ are almost certainly not the same.]

Answer 1

In what follows I will assume that $G$ is discrete and that $X$ is a simplicial complex with regular $G$-action (see Bredon's "Introduction to compact transformation groups", Chapter III.1). The regularity condition can be ensured by passing to the second barycentric subdivision if necessary. Under these conditions (unless I have overlooked something), the invariant cochains on $X$ are the same thing as cochains on the orbit space $X/G$.

A $k$-cochain on $X/G$ is a function $\phi$ from the $k$-simplices of $X/G$ to $A$. Let $\pi:X\to X/G$ denote the orbit projection. Then $\pi^*\phi$, which assigns to a simplex $\sigma\subseteq X$ the value $\phi(\pi(\sigma))$, is an invariant $k$-cochain on $X$. (That $\pi(\sigma)$ is a simplex of $X/G$ is ensured by regularity.)

Conversely, given an invariant $k$-cochain $\tilde\phi$ on $X$, we obtain a cochain $\phi$ on $X/G$ whose value on a simplex $\sigma\subseteq X/G$ is given by $\tilde\phi(\tilde\sigma)$, where $\tilde\sigma\subseteq X$ is any lift of $\sigma$. Regularity ensures that these lifts all differ by an element of $G$, so that this is well-defined.

Hence under these (topologically quite mild) conditions your invariant cohomology $H^*(\mathcal{C}_G)$ is the orbit space cohomology $H^*(X/G)$. You can then interpret your homomorphism $H^*(\mathcal{C}_G)\to H^*_G(X;A)$ as being induced by the map $$X_G:=EG\times_G X\to X/G$$ which collapses $EG$ to a point. When the action is free, this is a (weak) homotopy equivalence, and so $H^*(\mathcal{C}_G)\to H^*_G(X;A)$ is an isomorphism. In general, this map can be analysed using the Leray spectral sequence (see Borel's "Seminar on transformation groups", for example).

If you are interested in $G=\mathbb{Z}/p$ or $G=S^1$, you should look up "Smith-Richardson theory".

Answer 2

If $G$ is a finite group acting by simplicial automorphisms on a simplicial complex $X$, then

$$H^{\ast}(X/G;\mathbb{Q}) = (H^{\ast}(X;\mathbb{Q}))^{G} = H^{\ast}((C^{\ast}(X;\mathbb{Q}))^G),$$

where the superscripts indicate that we are taking the invariants with respect to $G$. For a proof, see Proposition 1.1 (and its proof) in my note "The action on homology of finite groups of automorphisms of surfaces and graphs", available on my webpage here. It is definitely necessary to take $\mathbb{Q}$ coefficients here (the referenced proof will make it clear why this is the case).