group cohomology and singular cohomology for $K(G,1)$

Question

For $K(G,1)$ space, I know that there is an isomorphism between group cohomology and singular cohomology. Is there an example for which this fact is useful (for example, it is hard to calculate group cohomology but easy to calculate singular cohomology).

Answer 1

I mean... Off the top of your head do you see what the cohomology of $\mathbb{Z}$ should be? Now ask yourself if you can see immediately what the cohomology of $S^1$ is. For me, at least, one of these questions is much easier. More broadly, what's the cohomology of a free group $F_k$? What about a wedge of $k$ circles?

These examples are slightly silly because they're one dimensional, but the idea is still there:

$K(G,1)$ spaces (when we can get a handle on what they look like) let us leverage our geometric intuition to solve group cohomological problems. I would argue that $K(\mathbb{Z}/2, 1) = \mathbb{R}P^\infty$ is a similar example (which is less silly, seeing as it's infinite dimensional). It admits a famous cellular decomposition, from which is cohomology follows immediately (especially over $\mathbb{Z}/2$ coefficients!).

I find that much easier to conceptualize and compute than working with a projective resolution of $\mathbb{Z} / 2$ directly.

I'll agree with you, though. Perhaps it's because I'm not an algebraic topologist, but I don't often find examples "in the wild" where $K(G,1)$ is a much simpler thing to consider than $G$ itself. When it happens, though (as in the cases above), it's really convenient.

Group cohomology is, in general, a rather complicated thing. After all, it encodes the solution to the extension problem in $H^2$! So it might be asking a bit much for $K(G,1)$ to be substantially simpler for a random group $G$.

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I hope this helps ^_^