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  1. Is there a low level "homotopical" description of cohomology with local coefficients? Similar to the identification of ordinary singular cohomology with the homotopy classes of maps to the Eilenberg-MacLane space $K(G,n)$.

I saw the page in nLab, but it's written in the language I don't understand. I also found this more general question with some nice clarifications by Urs.

I'm guessing that it means the following. Given an action of $\pi_1(X)$ on an (abelian) group $G$, we get a bundle over $X$ with fiber $G$, from which we get a bundle over $X$ with fiber $K(G,n)$ (or whatever its nonabelian analog). Then $H^n(X;G)$ is the homotopy classes of sections of the latter bundle. Is that interpretation correct?

  1. Is there a geometric object we can associate to the cohomology in the case of $n=1$? Similarly how we can use an element of untwisted cohomology to make a $G$-bundle, is it possible by taking an element of twisted cohomology to glue some kind of space (perhaps not from trivial pieces)?

Of course, I would be interested if the second question has an answer for $n>1$, but I don't even know what it is for the untwisted theory.

Edit: As was suggested in comments, I looked at the paper by Gitler. He constructs the following space $L_{\pi}(G,n) = E\pi \times _{\pi} K(G,n)$ (the quotient space under the diagonal action), where $\pi = \pi_1(X)$. Then we can get $H^n(X;G)$ as homotopy classes of maps $X$ to $L_{\pi}(G,n)$ that induce the identity on $\pi_1$.

It seems that we can pullback the bundle $E\pi \times _{\pi} K(G,n) \to B\pi$ using the classifying map of the universal cover $X \to B\pi$ to obtain a bundle with fiber $K(G,n)$ over $X$. Then if we have a map $X \to L_{\pi}(G,n)$ that induce the identity on $\pi_1$, we can turn it into a section of that bundle. Thus, it seems to confirm my guess for an answer to question 1.

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    $\begingroup$ See the question mathoverflow.net/q/331978/8103 and its answer. Does this answer your questions? $\endgroup$
    – Mark Grant
    Commented Jul 18, 2022 at 17:24
  • $\begingroup$ @MarkGrant That's very interesting, I will try to understand it better. One difference is that I don't want my coefficients to be a vector space necessarily. $\endgroup$
    – mathquest
    Commented Jul 18, 2022 at 20:52
  • $\begingroup$ There is no requirement that the coefficients be vector spaces in the paper by Gitler referenced in my answer. $\endgroup$
    – Mark Grant
    Commented Jul 18, 2022 at 21:02
  • $\begingroup$ Good, I will look at that paper. It's probably not hard to see that it's the same construction as described by homotopy theorists that I don't quite understand. $\endgroup$
    – mathquest
    Commented Jul 19, 2022 at 14:29

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