6
$\begingroup$

Let $C$ be a site, $\mathbf{S}$ some ($\infty$-? homotopy?) category of spaces.

Question. What do you call a (covariant!) functor $F:C\to \mathbf{S}$ enjoying the following property: for every hypercovering $a\colon V_\bullet \to U$ in $C$, the induced map $$ {\rm hocolim}\, F(V_\bullet)\longrightarrow F(U) $$ is a homotopy equivalence?

Have such things been studied? References are welcome.

$\endgroup$

1 Answer 1

6
$\begingroup$

This is just a sheaf valued in the ∞-category $\mathrm{Space}^{op}$. It is usually called a cosheaf. A place where this kind of thing shows up is in factorization algebras, that can be described as particular cosheaves over the Ran space.

I don't think they behave significantly differently from sheaves in any other complete ∞-category, so I do not believe there are specific references for them.

$\endgroup$
4
  • $\begingroup$ Thank you! I'll digest this and see if I have any follow-up questions... $\endgroup$ Feb 22, 2018 at 9:55
  • $\begingroup$ P.S. I am surprised that this doesn't seem to be a classical notion, since such functors appear in nature quite often. For example, if $f:X\to Y$ is a sufficiently nice map of topological spaces, then $(U\subseteq Y) \mapsto f^{-1}(U)$ should be such a "homotopy cosheaf" on $Y$. $\endgroup$ Feb 22, 2018 at 18:24
  • $\begingroup$ @PiotrAchinger Maybe I was unclear. This is a classical notion, it is just not significantly different from sheaves with values in any other (∞-)category. I am a bit uncertain of what kind of results you expect, do you have some statement in mind? $\endgroup$ Feb 22, 2018 at 18:34
  • $\begingroup$ For example: you can give exactly the same definition for sets: a cosheaf of sets is just a sheaf valued in $\mathrm{Set}^{op}$. These things do occasionally appear, but they do not seem to me to be special in any way. Cosheaves valued in stable ∞-categories or abelian categories are more interesting: for example they are the natural setting for Verdier duality (since sections of a sheaf with proper support form naturally a cosheaf). $\endgroup$ Feb 22, 2018 at 18:44

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.