This question loosely elaborates on an earlier question. It is pretty silly, but I'd like to hear some authoritative answers.

Recall that if $f:S^{\bullet}\to T^{\bullet}$ is a quasi-isomorphism of sheaves over $X$, which is, say, a manifold, then for every open set $U\subset X$, we have an induced isomorphism $R\Gamma(U,S^{\bullet})\to R\Gamma(U,T^{\bullet})$, so $H^i(U,S^{\bullet})\cong H^i(U,T^{\bullet})$ and in particular $H^i(X,S^{\bullet})\cong H^i(X,T^{\bullet})$.

To what extent is the converse true? At the coarsest level, when does a canonical isomorphism $R\Gamma(X,S^{\bullet})\to R\Gamma(X,T^{\bullet})$ reflect an underlying derived equivalence?

For a counterexample to the coarsest case, I believe the following serves: Consider a space $X$. Consider the constant sheaf on $k_X$. Let $f:X\to x_0$ be the retraction to a point $x_0\in X$. By standard theorems, we know that $H^i(X,Rf_*k_X)\cong H^i(X,k_X)$, but evaluating $Rf_*k_X(U)$ on any open subset $U$ missing $x_0$ assigns zero, as the fiber is empty. So in general these sheaves are not derived equivalent. What if $X$ deformation retracts to $x_0$? Is $k_X$ and $Rf_*k_X$ derived equivalent then? What if the homotopy doesn't have some Vietoris-Begle type behavior? See Kashiwara Schapira 2.7.8.

  • $\begingroup$ A comment about terminology: usually "derived equivalence" refers to a triangulated equivalence between derived categories. I think what you are calling "derived equivalence" above is usually called "quasi-isomorphism" or just "isomorphism". $\endgroup$ – Geordie Williamson Oct 23 '11 at 12:58
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    $\begingroup$ I realize that, but for someone who is more familiar with algebraic topology, where quasi-isomorphism refers to inducing isomorphisms on cohomology groups of a space, i.e. sheaf cohomology of the constant sheaf, then one is tempted to think that quasi-iso means a map inducing isomorphisms on sheaf cohomology and not cohomology sheaves (although the presheaf description of the latter is close to the former). I have to admit that although the definition was unambiguous, my semantic web conflated my intuition of the two. Thanks to Sam's answer, we now know when the two agree. $\endgroup$ – Justin Curry Oct 23 '11 at 15:28

I think you are asking: when is the functor $R\Gamma$ conservative (in the derived sense - i.e. if $R\Gamma (f)$ is a quasi-isomorphism then $f$ is a quasi-isomorphism). This is equivalent to $R\Gamma$ having no kernel - i.e. if $R\Gamma (F) \cong 0$, then F $\cong 0$ (by taking cones).

If you restrict to the triangulated subcategory generated by the constant sheaf $k_X$, I claim that $R\Gamma$ is conservative. This category can be thought of as something like derived unipotent local systems on $X$. The Barr-Beck theorem says that it is equivalent to dg modules over the algebra $C^\ast (X)$ of cochains on $X$ (so it is still interesting even if $X$ is simply connected).

Proof of claim: Let $f : X \to pt$ be the projection, and $C$ a (complex of) sheaves on $X$ such that $R\Gamma (C) = Hom _X(k_X , C) \cong 0$. Then if $T$ is any element of the triangulated subcategory generated by $k_X$, $Hom_X (T,C) \cong 0$. In particular, if $R\Gamma (T) \cong 0$, then $Hom _X(T,T) \cong 0$, so $T\cong 0$.

I think that this should be the largest triangulated subcategory of sheaves on $X$ on which global sections are conservative, but I am not sure how to prove this in general. It would follow from the Barr-Beck theorem if we knew that $R\Gamma$ preserved geometric realizations of simplicial objects, for example.

This situation is reminiscent of quasi-coherent sheaves on an affine scheme - this category is generated by $\mathcal O_X = f^\ast k$, and is equivalent to modules over $\Gamma (\mathcal O_X)$.

About your question: if $X$ deformation retracts to a point, then are $k_X$ and $Rf_\ast k_X$ quasi-isomorphic? (where by $Rf_\ast k_X$ I assume you mean the corresponding skyscraper sheaf on $X$). Doesn't your argument just above work (provided $X$ is not a point)?

  • $\begingroup$ This is an excellent answer. Thank you. $\endgroup$ – Justin Curry Oct 23 '11 at 3:15

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