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Let $M$ be a finite CW-complex. Let $F$ be a finite rank local system over $M$ with coefficients in any field. Is it true that $\dim(H^k(M,F))$ is at most the number of $k$-cells times $\operatorname{rank}(F)$?

If $F$ is the trivial local system then this result is proven in almost any standard textbook in topology (or at least immediately follows from there). I believe that it should be true in the above generality and would be happy to have a reference.

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I'm not sure of a reference off the top of my head, but the proof is straightforward. Let's say the number of $k$-cells of $M$ is $n_k$.

Because $M$ is given a CW structure, its universal cover $\tilde M$ is given a $\pi_1(M)$-invariant CW structure; because $\pi_1(M)$ acts freely, the cellular chain complex $C_*(\tilde M;R)$ is a free $R[\pi_1(M)]$-module equipped with a canonical isomorphism $C_*(\tilde M;R) \otimes_{R[\pi_1(M)]} R \cong C_*(M;R)$. In particular, $C_k(\tilde M;R)$ is a free $R[\pi_1(M)]$-module of rank $n_k$.

Suppose we are given a local system $\rho: \pi_1(M) \to \text{Aut}_R(A)$, where $A$ is a finitely generated module over some Noetherian ring $R$. The cellular chain complex with local system $\rho$ is given by $$C_*(M;\rho) = C_*(\tilde M;R) \otimes_{R[\pi_1(M)]} A,$$ with $\pi_1(M)$ acting on $A$ by $\rho$.

Then each $C_k(M;\rho)$ is isomorphic to $A^{n_k}$, which is finitely generated; it has a submodule $Z_k \subset A^{n_k}$ of $k$-cycles, and the homology group $H_k(M;\rho)$ is the quotient $Z_k/B_k$, for the submodule $B_k \subset Z_k$ of boundaries.

Every submodule of a finitely generated module over a Noetherian ring is finitely generated, and so $Z_k$ is finitely generated, and hence so is its quotient $H_k(M;\rho)$.

This level of generality includes $A$ any finite-dimensional vector space over a field and $A$ any finitely generated abelian group.

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    $\begingroup$ The punchline is that there is an isomorphism of graded $R$-modules $C_*(M;\rho) \cong C_*(M;A)$; you're just twisting the differential. So whatever argument you would use to show that $H_*(M;A)$ is finitely generated will apply to show that $H_*(M;\rho)$ is, as well. $\endgroup$
    – mme
    Commented May 30, 2019 at 0:48

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