Does taking homology commute with taking coinvariants? Suppose we are given a chain complex of vector spaces $(C,d)$ over a field (of possibly not char=0) and in each degree a representation $G_i$ acting on $C_i$. However, I know that the chain complex descends to a well defined chain complex $(C_G,d)$ where for a vector space $V$ I denote $V_G=V/<v-g\cdot v, g\in G>$ the coinvariants.
Thus I have now two chain complexes and two homologies to compare. When is it true that $H(C,d)_G=H(C_G,d)$? Or can you at least give me a hint what theories I have to look for to get a grip on such a situation?
I heard about group cohomology but I am not sure if this is what I am looking for.
 A: (Co)homology commutes with exact functors, but taking coinvariants is only right exact and not left exact in general. Suppose we start with a short exact sequence of $G$-equivariant maps
$$V_\bullet\colon \quad 0 \to V_2 \to V_1 \to V_0 \to 0$$
We have then $H_\bullet (V_\bullet) = 0$, as this thing is exact. Now the coinvariant functor $V \rightsquigarrow V_G$ (which is usually denoted by subscript $G$, as superscript $G$ is reserved for invariants, i.e. fixed points) is right exact but not left exact, so in general you obtain an exact sequence
$$(V_\bullet)_G\colon \quad (V_2)_G \to (V_1)_G \to (V_0)_G \to 0$$
and
$$H_2 ((V_\bullet)_G) = \ker ((V_2)_G \to (V_1)_G) \ne 0.$$
A silly example: let the cyclic group $C_2$ act on $k^2$ by permuting the coordinates. We have a $C_2$-equivariant subspace given by the line $L$ spanned by $(1,-1)$, and the inclusion $L \rightarrowtail k^2$ induces a trivial map when we pass to co-invariants.

This is indeed related to group homology (not co-homology). What we have is a long exact sequence
$$\cdots \to H_1 (V_2, G) \to H_1 (V_1, G) \to H_1 (V_0, G) \xrightarrow{\delta_1} (V_2)_G \to (V_1)_G \to (V_0)_G \to 0$$
—this follows from the definition of group homology $H_n (-, G) = L_n (-)_G$ as the left derived functors of the right exact functor of coinvariants. So we have
$$H_2 ((V_\bullet)_G) = \ker ((V_2)_G \to (V_1)_G) = \operatorname{im} (H_1 (V_0, G) \xrightarrow{\delta_1} (V_2)_G).$$
