# Sheaves with isomorphic cohomology, but not quasi-isomorphic

Suppose I have two (constructible) sheaves of vector spaces $F$ and $G$ over the same base space that have isomorphic cohomology (degree by degree), but no sheaf map inducing this isomorphism (i.e. they are not quasi-isomorphic).

Now imagine I apply any of Grothendieck's 6 operations (or other functors) to $F$ and $G$, will the resulting sheaves, say $\Psi(F)$ and $\Psi(G)$, have isomorphic cohomology as well?

Thanks!

EDIT: I suppose the answer in general is no. Consider an injective (acyclic) sheaf and any functor that doesn't preserve injectives. A concrete example would be nice though.

EDIT 2: Thanks to Algori's comments I need to substantially limit the sheaves of interest. Assume additionally that the sheaves have identical support and furthermore that when we take the function that assigns to a point the euler characteristic of the stalk over that point (thereby producing a constructible function) to these two sheaves, these functions are identical. If we are using real sub-analytic partitions then a theorem of Kashiwara's says that the Grothendieck group of the bounded derived category of real constructible sheaves is isomorphic to the group of constructible functions. If I understand things correctly this would mean that these two sheaves are equivalent in the Grothendieck group (which seems weird since this function ignores gluing data). I am dealing with particular examples of two constructible (cellular) sheaves over a simplicial complex, which are not isomorphic yet produce identical constructible functions.

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This is a trivial example, but still: consider a space with more than one closed point and the constant sheaves supported on two different closed points. – algori May 26 '11 at 20:20
.. or did you mean to ask about maps in the derived category? – algori May 26 '11 at 20:28
(.. in which case the same argument applies) – algori May 26 '11 at 20:46
@algori, What is the functor that you are applying? Yes, maps in the derived category, but other examples are good. – Justin Curry May 26 '11 at 21:59
Justin -- e.g. the pullback to one of the points. – algori May 26 '11 at 22:58

Let $X$ be the wedge of two circles. To give a local system on $X$ with stalk a complex vector space $V$ is to give two matrices $A_1,A_2\in GL(V)$. If the subgroup $\langle A_1,A_2\rangle$ of $GL(V)$ generated by $A_1,A_2$ has no invariants in $V$, then $H^0$ of the resulting local system will be $0$ and $H^1$ will be of rank $\dim V$ for Euler characteristic reasons.
To give a map in $Sh(X)$ between the local systems that correspond to $(V,A_1,A_2)$ and $(W,B_1,B_2)$ is to give a linear map $f:V\to W$ such that $f A_1=B_1f,f A_2=B_2f$. Now suppose that $V=W$, that both the groups $\langle A_1,A_2\rangle$ and $\langle B_1,B_2\rangle$ have no invariants in $V$, that $A_1$ is diagonalizable and that the eigenvalues of $A_1$ and $B_1$ are pairwise distinct. Then any map between the local systems is zero.
Justin -- I think so. E.g., tensor both with the dual of one of them. Then in one case there will be a constant subsheaf, hence $H^0$ will bo $\neq 0$, while in the other case probably not. – algori May 27 '11 at 22:52