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Whenever we have a continuous map of topological spaces $f: X \to Y$ and a sheaf $\mathcal{F}$ on $Y$ (of abelian groups for example), we get an induced pullback map $$ H^n(Y, \mathcal{F}) \to H^n(X, f^* \mathcal{F}), $$ which is described for example in Iversen, Cohomology of Sheaves, II.5.1. I would like to have an explicit description of this map in one particular case.

Suppose $X = Y = S^1$ and further suppose that the sheaf is a local system $\mathcal{L}$. Then its pullback $f^* \mathcal{L}$ is also a local system. We have a nice description of local systems in terms of representations of the fundamental group, and the pullback is easily describable in this language: if $$ \pi_1(Y) \to \mathrm{Aut}(\mathcal{L}_t) $$ is the representation defining $\mathcal{L}$, then $f^* \mathcal{L}$ is described by the composition $$ \pi_1(X) \xrightarrow{f_*} \pi_1(Y) \to \mathrm{Aut}(\mathcal{L}_t). $$

The cohomology groups of a local system on $S^1$ are also nicely described in terms of representations: if we denote by $T \in \mathrm{Aut}(\mathcal{L}_t)$ the image of a generator of $\pi_1 (S^1)$, then $$ H^0(S^1, \mathcal{L}) \cong \mathrm{Ker}(T - id) $$ $$ H^1(S^1, \mathcal{L}) \cong \mathrm{Coker}(T - id) $$ $$ H^n(S^1, \mathcal{L}) = 0), \quad \text{for any } n>1 $$ This is proved using Cech cohomology.

Now the homotopy type of our map $f: S^1 \to S^1$ is determined by its degree $d$, and the automorphism of the stalk that defines $f^* \mathcal{L}$ is just $T^d$. Since $\mathrm{Ker}(T - id) \subset \mathrm{Ker}(T^d - id)$ it seems tempting to say that the pullback map on $H^0$ is the inclusion. However I don't know how to prove this, because I haven't found any result about some possible functoriality of Cech cohomology with respect to the covering. Furthermore, I don't know what the map on $H^1$ could look like. The first candidate I can think of would be some projection between the cokernels, but $\mathrm{Im}(T - id) \subset \mathrm{Im}(T^d - id)$ seems unlikely because of the first isomorphism theorem (if the kernels get larger then the images should get smaller).

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The circle $S^1$ is a $K(\mathbb{Z},1)$ space, so the cohomology of a local system on it can be identified with group cohomology. Let $R=\mathbb{Z}[T,T^{-1}]$ be the group ring of $\mathbb{Z}$. Given a local system $\mathcal{L}$, $$H^i(S^1,\mathcal{L})= Ext_R^i(\mathbb{Z}, \mathcal{L})$$ This can be computed as using the formulas you wrote, because $$0\to R\xrightarrow{T-1}R\to \mathbb{Z}\to 0$$ is free resolution.

The degree $d$ map on the circle corresponds to $f:R\to R$ by $T\mapsto T^d$. This determines a map on group cohomology. This does not seem compatible with the above resolution, bu you can compute it in other ways. Using derivations mod inner derivations for $H^1$ is probably better for what you want.

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    $\begingroup$ To make this compatible with that resolution, you can map by multiplication by $1 + T + T^2 + \dots + T^{d-1}$ on the earlier copy of $R$. I think this shows that the multiplication-by-$1 + T + T^2 + \dots + T^{d-1}$ map on coinvariants is the map on $H^1$. $\endgroup$
    – Will Sawin
    Commented Oct 13, 2021 at 18:31
  • $\begingroup$ This seems really useful! I didn't know about this equivalence at all. Do you know of a reference where I can read about this? Eilenberg-MacLane spaces, group cohomology and the equivalence. For instance, if I'm actually interested in local systems of Q-vector spaces how does this affect the equivalence? $\endgroup$ Commented Oct 14, 2021 at 8:57
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    $\begingroup$ @EduardodeLorenzo One reference which I like is Brown Cohomology of groups. Everything goes through with $\mathbb{Q}$-coefficients. $\endgroup$ Commented Oct 14, 2021 at 12:40

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