13
$\begingroup$

Suppose that we are given a nice space $X$ and a sheaf of abelian groups $F$ on $X$. Fix an integer $n$. Then We have a contravariant functor from nice spaces over $X$ to abelian groups; Namely, to a space $f: Y \to X$ we associate the abelian group $H^n (Y, f^{*}F)$ (Sheaf cohomology).

If $X$ is a point, Then this functor is represented, in the homotopy category, by the Eilenberg-Maclane space $K(F,n)$.

My questions are:

1) Can one formalize what will it mean for our functor to be representable, "homotopically"? I am not very sure, but I suspect that the most naive definition of homotopy category of spaces over $X$, and the requirement that the functor is representable in this category, are not right (and I did not check that the functor actually factors to this "homotopy" category).

2) Is it representable?

$\endgroup$

1 Answer 1

8
$\begingroup$

Sheaf cohomology over $X$ is representable in the homotopy category of oo-stacks over X / spaces over X, yes.

Details, links and references are at http://ncatlab.org/nlab/show/cohomology

$\endgroup$
2
  • 6
    $\begingroup$ Urs -- maybe it's just me but I've taken a look at that page and I have no idea what the theorem is. Could you please state it explicitly: for any $Y$ over $X$ we have $H^{\ast}(Y,f^{-1}F)$ is the $Hom$ in the category such and such from $Y$ to the object such and such constructed from $X$ and $F$? $\endgroup$
    – algori
    Jul 20, 2010 at 23:47
  • 2
    $\begingroup$ The space Y itself is a sheaf over X, albeit not a sheaf of abelian groups, but a "nonabelian sheaf". You regard both Y as well as F as being oo-stacks over X (sheaves of simplicial sets, essentially): Y is just a sheaf of sets. for F you take the simplicial presheaf that you obtain under the Dold-Kan map from the chain complex of sheaves concentrated on F in the dired degree n. Call that B^n F. Then the cohomology in question is [Y,B^n F] in the homotopy category of the oo-stack oo-category. See the old article by Kenneth Brown that is linked at the entry. $\endgroup$ Jul 21, 2010 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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