Suppose we have a homotopy equivalence $f: X \to Y$ (with homotopy inverse $g: Y \to X$) and a local system (i.e. a locally constant sheaf) $\mathscr{S}$ on $Y$. Is the homomorphism
$$f^*: H^k(Y,\mathscr{S}) \to H^k(X, f^{-1}\mathscr{S})$$
induced by the unit $1 \Rightarrow f_*f^{-1}$ of the adjunction then an isomorphism?
I have a feeling that this should be the case. Under the equivalence between locally constant sheaves and representations of the fundamental groupoid (on sufficiently nice spaces), we have that $(f \circ g)^{-1}\mathscr{S} \cong \mathscr{S}$ and $(g \circ f)^{-1}\mathscr{L} \cong \mathscr{L}$ for $\mathscr{L}$ a locally constant sheaf on $X$. I'm pretty sure this part is correct.
Thus, from the unit map $f^*\mathscr{S} \to g_*g^{-1}f^{-1} \mathscr{S} \cong g_*\mathscr{S}$ we get a morphism of sheaf cohomology groups $$g^*: H^k(X, f^*\mathscr{S}) \to H^k(Y, g^{-1}f^{-1}\mathscr{S}) \cong H^k(Y, \mathscr{S}).$$ I think that this should be the inverse to the morphism $f^*$ above.
The usual way to show this would be first prove the following Lemma:
Lemma Given a locally constant sheaf $\mathscr{S}$ on $Y$ and homotopic maps $f_0: X \to Y$ and $f_1: X \to Y$, then the induced maps in cohomology are equal: $$f_0^*=f_1^*: H^k(Y,\mathscr{S}) \to H^k(X, f^{-1}\mathscr{S}).$$
Unfortunately, I'm not very experience with sheaves yet, so I'm not sure how to get this result.
For motivation, the particular application I have in mind is for cohomology with local coefficients on a manifold with boundary, where the coefficients are given by the sheaf of parallel sections of a vector bundle with flat connection. I know a manifold with boundary is homotopy equivalent to it's interior, so if we have a vector bundle with flat connection on the manifold with boundary, is the twisted de Rham cohomology the same as the twisted de Rham cohomology when we pull the flat vector bundle back to the interior?