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Consider the co-presheaf $\mathcal{F}$ of continous real-valued functions with relatively-compact support on a topological space $X$. Consider a point $x\in X$.

1) When $\mathcal{F}$ is considered a co-presheaf with values in the category of sets, what is the co-stalk $\mathcal{F}_x$?

2) For X paracompact a partition of unit exists. Hence $\mathcal{F}$ is even a cosheaf when considered with values in the category of real vector spaces. What is the co-stalk $\mathcal{F}_x?$

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  • $\begingroup$ I had trouble understanding this at first. But I guess $\mathcal{F}(U)$ is by definition the set of continuous functions whose support is contained in $U$? $\endgroup$ Commented Mar 7, 2016 at 18:12
  • $\begingroup$ @Todd Trimble I made an edit: The closure of the support is a compact subset of U. $\endgroup$
    – Jo Wehler
    Commented Mar 7, 2016 at 18:20

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Projective limits in vector spaces and in sets are the same so the stalk does not depend on whether you consider this as a co-presheaf of sets or vector spaces.

in both case it is just the directed projective limit of the $\mathcal{F}(U)$ for $U$ among neighbourhood of $x$, the corestriction maps $\mathcal{F}(U) \rightarrow \mathcal{F}(U')$ are all injective, so this projective limit is just an intersection: the stalk at $x$ is hence the set of functions on $X$ whose support is included in all neighbourhood of $x$... i.e. it is zero most of the time (unless the closure of $\{x\}$ is an open sets in which case it is $\mathbb{R}$).

At an informal level: sheaves can be thought of as functions and cosheaves as measures (you can integrate a sheaf against a cosheaf using a coend, multiply a cosheaf by a sheaf to get a cosheaf etc...). Functions tend to be determined by values at points (i.e. stalks) but for measure the "value at a point" is something like $\mu(\{x\})$ and it is far from being enough to understand the measure

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  • $\begingroup$ As I understand: In case of X an open subset of R^n the stalks are zero. - Can you name a reference concerning the heuristics "co-sheaves as measures", thanks. $\endgroup$
    – Jo Wehler
    Commented Mar 7, 2016 at 18:26
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    $\begingroup$ Indeed (when n>0): if the space is hausdorff the only points where the stalk is non-zero are the isolated points. For the reference, I will try to find something. But roughly if you have a co-complete monoidal closed category $C$ (so sets, abelian group, vector spaces etc...) you can show that there is an equivalence of category between cosheaves with values in $C$ and co-limit preserving $C$-enriched functor from $Sh(X,C)$ to $C$. So if you think as $C$- valued sheaves as "functions with values in $C$", cosheaves are the things that allows to integrate those functions. $\endgroup$ Commented Mar 7, 2016 at 18:40
  • $\begingroup$ +1: nice answer, I feel I almost understand Cosheaves now ...! $\endgroup$ Commented Feb 13, 2019 at 3:34

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