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?

  • 6
    $\begingroup$ 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. $\endgroup$ Feb 21, 2013 at 17:11
  • $\begingroup$ +1 for banana sheafs :-D $\endgroup$ Feb 21, 2013 at 20:47
  • 2
    $\begingroup$ I didn't know that the category of bananas had kernels. $\endgroup$ Feb 20, 2014 at 2:56

1 Answer 1


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 $S$.

  • 4
    $\begingroup$ 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. $\endgroup$
    – Jacob Bell
    Feb 21, 2013 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.