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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?

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    $\begingroup$ The fact that $f$ and $g$ are homotopy inverse implies that $g_* \mathcal V = f^* \mathcal V$. The first two are true by the exact same proof as that cohomology itself is a homotopy invariant. I don't know what shriek is so can't comment on the last two. $\endgroup$
    – user98602
    Commented May 26, 2016 at 15:18
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    $\begingroup$ The shriek doesn't even preserve local constancy, right? Why insist on local coefficients rather than sheaf cohomology? Anyway, in your case the inclusion is proper so shriek and star coincide. $\endgroup$ Commented May 26, 2016 at 15:51
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    $\begingroup$ @MikeMiller No that is not true, if $g:\{0\}\rightarrow\mathbb{R}^n$ is the inclusion of the origin, $g_*\mathcal{V}$ is supported in $\{0\}$ so it is not a local system (whereas $f^*\mathcal{V}$ is) $\endgroup$
    – Roland
    Commented May 26, 2016 at 19:06
  • $\begingroup$ @Roland I just saw this a couple months late. Thanks for the correction. $\endgroup$
    – user98602
    Commented Aug 30, 2016 at 4:25
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    $\begingroup$ It suffices to see that homotopic maps induce the same homomorphism, which is basically the same as the standard proof. To avoid slight annoyances like identifying $f_0^*\mathcal S$ and $f_1^*\mathcal S$ you could instead work with local coefficients as representations of $\pi_1$. $\endgroup$
    – user98602
    Commented Sep 3, 2016 at 15:24

2 Answers 2

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I succeeded in proving it myself:

Suppose we have homotopy inverses $f: X \to Y$ and $g: Y \to X$, where $X$ and $Y$ are locally connected and semi-locally simply connected (e.g. manifolds). Let $\mathcal{C}$ be a local system (a locally constant sheaf) over $Y$.

The adjoint equivalence of categories between sheaves and etale spaces restricts to an adjoint equivalence between local systems and covering spaces (this works for sheaves of sets, groups, modules etc.)

Then by homotopy invariance of fibre bundle pullbacks (on nice spaces), $f^*$ and $g^*$ will be homotopy inverses (i.e. giving an equivalence of categories) between the categories of covering spaces over $X$ and $Y$, and hence between the corresponding categories of local systems of these two spaces.

Without loss of generality, lets take $g^*$ to be the right adjoint to $f^*$ (since an equivalence of categories can be promoted to an adjoint equivalence).

Then using the natural isomorphisms $\Gamma(Y, \mathcal{C}) \cong \mathrm{Hom}(\mathbb{1}, \mathcal{C})$ and $f^{-1} \mathbf{1} \cong \mathbf{1}$ where $\mathbf{1}$ is the tensor unit (e.g. the constant sheaf of integers if we work with sheaves of abelian groups), we get natural isomorphisms $$\Gamma(Y, \mathcal{C}) \cong \mathrm{Hom}(\mathbf{1}, \mathcal{C}) \cong \mathrm{Hom}(\mathbf{1}, g^{-1}f^{-1}\mathcal{C}) \cong \mathrm{Hom}(f^{-1}\mathbf{1}, f^{-1}\mathcal{C}) \cong \mathrm{Hom}(\mathbf{1}, f^{-1}\mathcal{C}) \cong \Gamma(X, f^{-1}\mathcal{C}).$$

The (homology of the) right derived functor $R\Gamma(X, -)$ is sheaf cohomology, so we are done.

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I don't think that your argument is correct. Proving that two functors are isomorphic when restricted to the full subcategory of locally constant sheaves is not enough to deduce an isomorphism between their derived functors, because the definition of derived functor uses the whole category (locally constant sheaves are almost never acyclic).

For a proof (at least in the constant case) see Kashiwara-Shapira, Sheaves on Manifolds, Proposition 2.7.5. Essentially they prove a vanishing theorem for cohomology on $[0,1]$ and use proper base change for the projection $X \times [0,1] \rightarrow X$ to deduce that homotopic maps induce the same morphisms in cohomology for constant sheaves.

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