# A Comparison between $\pi_{1}$ of cohomology and cohomology of $\pi_{1}$

Let we have a complex of abelian topological or lie groups $$\ldots \to G_{n}\to G_{n+1}\to \ldots$$ such that the image of $G_{n}$ is a closed subgroup of $G_{n+1}$. Then we have a complex of fundamental groups $$\ldots \to \pi_{1}(G_{n})\to \pi_{1}(G_{n+1})\to \ldots$$

Are there some standard theorems or results about a comparison betwee the cohomology of $\pi_{1}(G_{n})$ and the $\pi_{1}$ of cohomolgy of $G_{n}$?

If you consider an exact sequence of groups, it's a fibration of topological spaces, so you get a homotopy long exact sequence. So that makes me think $\pi_n$ behaves like a derived functor from topological abelian groups to abelian groups. In particular this is the derived functor of $\pi_0$.
So there is the spectral sequence for applying a derived functor to a complex whose first page whose second page is $\pi_m$ of cohomology of $G_n$.
However, I don't think the first page is $\pi_m(G_n)$. One can see this from an exact sequence, like $\mathbb Z/2 \to S^1 \to S^1$. All the homotopy groups of the cohomology are trivial, because it is exact. But $\pi_m(G_n)$ has many nonzero terms and you can't make them all vanish just by applying a differential.
• thanks for the answer. I should concentrate on it. But another question: In this question, to what extent I need to consider "abelian" group. This is a motivation to ask:Let $H$ be a closed (but not necessarily normal) subgroup of a topological group $G$. Then is it true that the topological space $G/H$, left cosets, has an abelian fundamental group? – Ali Taghavi Nov 13 '14 at 22:54
• @AliTaghavi No, take a finite non-abelian subgroup of a connected topological group - say the quaternions inside $SU_2$. – Will Sawin Nov 13 '14 at 23:01