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Let $X$ and $Y$ be two topological spaces and $R$ be commutative ring. Let $D_c^b(X, R)$ and $D_c^b(Y,R)$ be their respective bounded derived categories of constructible sheaves of $R$-modules. I have three questions:

  1. If $f:X\rightarrow Y$ is a weak homotopy equivalence does it induce (under any of the standard cohomological operations) an equivalence between $D_c^b(X, R)$ and $D_c^b(Y,R)$?

  2. If 1. is not true in general, is there some interesting subcategory of topological spaces where it does become true? Perhaps CW-complexes?

  3. If 1. is not true does it become so if we restrict to the full subcategories of $D_c^b(X, R)$ and $D_c^b(Y,R)$ whose objects have local system cohomology?

I suppose what I'm really asking is whether it is possible to naturally associate a bounded derived categories of constructible sheaves of $R$-modules to a homotopy type?

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Regarding 3: the subcategories of locally constant objects are indeed equivalent provided that $X$ and $Y$ are very nice spaces like CW-complexes or paracompact topological manifolds. For such a space $X$, locally constant sheaves of bananas on $X$ are the same thing as local systems of bananas on the homotopy type of $X$. There's a proof in Appendix A.1 of "Higher Algebra" by J. Lurie. – Marc Hoyois Feb 21 '13 at 17:11
+1 for banana sheafs :-D – Johannes Hahn Feb 21 '13 at 20:47
I didn't know that the category of bananas had kernels. – Vidit Nanda Feb 20 '14 at 2:56

I don't know much about these things but I think 1. fails even if $X$ is a point and $Y$ is a line. Let's say also $R$ is a field. Then the category of constructible sheaves on $X$ is just $R$-Vect which is in particular semisimple, but the category of constructible sheaves on $Y$ is not semisimple: if we choose a sheaf on a point $p$ and on the complement $U$ of that point, then there are in general many non-isomorphic choices of a sheaf on $Y$ with given restrictions to $p$ and $U$ which fit in a short exact sequence.

So their derived categories are not equivalent either.

Addendum: Maybe I can say something about how the usual homotopy invariance of cohomology is visible in the 'six functor' formalism. For any space $X$ let $\pi_X$ be the projection to a point, then the cohomology of $X$ (with any coefficients) just corresponds to functor $R \pi_{X\ast} \circ \pi_X^\ast$. If $f \colon X \to Y$ is an arbitrary map, then note that $\pi_Y \circ f = \pi_X$ which implies $$ R\pi_{Y\ast} \circ Rf_\ast \circ f^\ast \circ \pi_Y^\ast = R \pi_{X\ast} \circ \pi_X^\ast;$$ and now the morphism $\mathbf 1 \to Rf_\ast \circ f^\ast$ coming from the adjunction gives a map $f^\# \colon R \pi_{Y\ast} \circ \pi_Y^\ast \to R \pi_{X\ast} \circ \pi_X^\ast$. Of course evaluating this equation on a choice of coefficients we get the usual map $H^\bullet(Y) \to H^\bullet(X)$. The correct way to express homotopy invariance is now that if $f$ and $g$ are homotopic maps $X \to Y$, then $f^\#$ and $g^\#$ are equal. This can be generalized to the relative situation, when $X$ and $Y$ are spaces over some base space $S$ and we consider the derived pushforward to $S$ instead of to a point, and we take homotopies over $S$.

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I'm not an expert either but it seems right. I guess the point is that constructible things have to do with stratifications, and stratifications are about the topology/geometry of a space, not its homotopy. – Jacob Bell Feb 21 '13 at 8:45

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