What is the sheafification of the presheaf of the one point compactification?

Question

Okay, so I had this idea for a presheaf that is quite peculiar. Instead of being based on algebraic category (i.e. abelian groups), it is based on a topological one, the category of compact topological spaces.

Okay, so let's say you have a space $X$. An open space $U$ is associated with the its one-point compactification, $U^*$. For each pair $V \subseteq U$, we have a continuous map $\text{res}_{V,U}: U^* \to V^*$ defined as

$$\text{res}_{V,U}(x) = \begin{cases} x & \text{if $x \in V$} \\ \infty & \text{if $x \in U \backslash V \vee x = \infty$} \end{cases}$$

$\text{res}_{U,U}=\text{id}_{U^*}$ and $\text{res}_{U,V} \circ \text{res}_{V,W} = \text{res}_{U,W}$ trivially. Therefore, this forms a presheaf.

My question is what is the sheafification of this presheaf? (I'm hoping it will be some sort of localized compactification!)

Answer 1

This is only a partial answer, but it's too long for a comment, so I'll post this as an answer and hope that nobody complains.

For simplicity, let's call your presheaf $\mathcal{F}$.

First, let's look at the stalk:

The stalk $\mathcal{F}_x$ is just going to be the direct limit (with respect to the restriction you defined) of the spaces $U^*$ with $U$ an open neighborhood of $x$. This is just going to be the space $$\mathcal{F}_x = \left(\bigcap_{U \ni x} U\right) \cup \{\infty\}$$ with the topology in which the only non-trivial open set is $\{\infty\}$.

A special case: the space is $T_1$

That $X$ is $T_1$ is equivalent to the intersection of the neighborhoods around $x$ being $\{x\}$ for every $x$. In this case, $\mathcal{F}_x = \{x\}^* = \{x,\infty\}$ with the topology $\{\emptyset,\{\infty\},\{x,\infty\}\}$. Recall that the sheafification $\mathcal{F}^{sh}(U)$ is a subset of $\prod_{x \in U} \mathcal{F}_x$, and in particular, since each $\mathcal{F}_x$ contains two distinguished points, this can be identified with the subsets of $U$, i.e. by looking at where this is not $\infty$. With this identification, $$\mathcal{F}^{sh}(U) = \{A \subseteq U \colon A~\mathrm{discrete}\}.$$ By a discrete space, I just mean every point is open. Restriction from $U$ to $V$ is just $A \mapsto A \cap V$.

Remark

Even for nice spaces, this sheafification will typically be pretty large. In particular, for our $T_1$ case, every finite subset is discrete!

Answer 2

This sheafification does not exist in general, at least if you want the target category to be the category of compact spaces. To construct the sheafification of a sheaf, you need the existence of certain limits and colimits, and the category of compact spaces does not have all limits and colimits (in particular, it does not have all equalizers and does not have any infinite colimit such that the colimit computed in the category of all spaces is not compact). For instance, suppose $X=\{0,1\}^\mathbb{N}$ is the Cantor set. A cofinal family of open covers of $X$ is given by $\mathcal{U}_n=\{\{x\}\times\{0,1\}^{[n,\infty)}:x\in\{0,1\}^n\}$, and all the sets in each $\mathcal{U}_n$ are disjoint. It follows that we should be able to compute the value of the sheafification on $X$ as the sequential colimit of the spaces $Y_n=\prod_{U\in \mathcal{U}_n} U^*$, with the maps $Y_n\to Y_{n+1}$ given by the obvious restrictions. But this colimit does not exist in the category of compact spaces (indeed, it is easy to see each map $Y_n\to Y_{n+1}$ is the inclusion of a proper clopen subset).

The example of the Cantor set was just used for the convenience of being able to explicitly write down a cofinal family of open covers. In general, the same problem will arise for almost every space: for instance, in any $T_1$ non-discrete space $X$, there is no single finest open cover of $X$, and so the colimit that would describe the value of the sheafification on $X$ probably does not exist.