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 $$\ldots \to \pi\_{1}(G\_{n})\to \pi\_{1}(G\_{n+1})\to \ldots$$



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> 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 $G\_{n}$?
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A Comparison between $\pi_{1}$ of cohomology and cohomology of $\pi_{1}$
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.


I don't know enough about spectral sequences to say exactly what the relationship between these two objects is, but I believe it is a question primarily of homological algebra.


