# Examples and definition of cocompact objects

An object $X$ of a locally small category $C$ that admits filtered colimits is called compact if $$\operatorname{Hom}_{C}(X,-)$$ preserves filtered colimits.

Let $C$ be a locally small category that admits filtered limits. Let's define an object $X$ of $C$ to be cocompact if $$\operatorname{Hom}_{C}(-,X)$$ sends cofiltered limits to filtered colimits in $Sets$.

I never heard of this definition before. Is it reasonable? If yes, what are typical examples of cocompact objects in $Sets$, $Top$, etc?

• You mean a cofiltered limit. Jul 17, 2014 at 0:44
• Cocompactness is not as useful as compactness because many categories of interest are $\mathbf{Ind}(\mathcal{A})$ for some $\mathcal{A}$, but not so many are $\mathbf{Pro}(\mathcal{A})$. Jul 17, 2014 at 7:09
• Thanks for the comments. Do you know of any examples or non-examples in $Sets$ or $Top$? Jul 17, 2014 at 9:52
• I believe the only cocompact objects in $\mathbf{Set}$ are $0$ and $1$. Jul 17, 2014 at 10:04

I prefer to call the quoted categorical concept of compactness strong compactness, since it is stronger than the usual notion of compactness for topological spaces.

We can express the definition of strong cocompactness of an object $$D$$ as the requirement that any morphism $$\lim_{i \in I}C_i \to D$$, with $$I$$ a cofiltered diagram, should factor through one of the limit projections $$\pi_j: \lim_{i \in I}C_i \to C_j$$. We see from this reformulation that if any subset/subspace of a set/space $$X$$ fails to be strongly cocompact, we can therefore deduce that $$X$$ is not strongly compact either; we use this to help us in our search.

In $$\mathbf{Set}$$, consider $$\prod_{n \in \mathbb{N}} 2$$ as a cofiltered limit of copies of finite products of the two-element set, $$2$$. Given any set $$X$$ with two or more distinct elements, pick two such elements and call them $$0$$ and $$1$$; label the elements of $$2$$ the same way. Then we have a function $$f: \prod_{n \in \mathbb{N}} 2 \to X$$ sending a sequence to $$0$$ if the sequence contains a $$1$$ and the first $$1$$ in the sequence is followed by a $$0$$; and to $$1$$ otherwise. By construction, this cannot factor through the projection from $$\prod_{n \in \mathbb{N}} 2$$ to any finite product. Clearly any morphism from a cofiltered limit to the one-element set $$1$$ factors through one (indeed any) of its projection maps. However, contrary to what Zhen Lin said in the comments, $$0$$ is not strongly cocompact in $$\mathbf{Set}$$, since if we consider the codirected limit of subsets $$[n,\infty) \subseteq \mathbb{N}$$, whose intersection is empty, we see that this is not preserved by $$\mathrm{Hom}(-,\emptyset)$$.

By passing across the forgetful-indiscrete adjunction $$(U \dashv I)$$ between $$\mathbf{Top}$$ and $$\mathbf{Set}$$, we can use the same argument as above to deduce that any indiscrete space with more than 2 points, or indeed any non-$$T_0$$ space, cannot be strongly cocompact in $$\mathbf{Top}$$. If the subspace of $$X$$ on the points $$0$$ and $$1$$ is homeomorphic to the Sierpinski space, such that $$0$$ is open, the argument still works, since the set $$f^{-1}(0)$$ of sequences described above is the union of opens $$\prod_{i=0}^{n-1} \{0\}_i \times \{1\}_n \times \{0\}_{n+1} \times \prod_{i=n+2}^{\infty} 2$$ in $$\prod_{n \in \mathbb{N}} 2$$. This union is disjoint and there can be arbitrarily many non-trivial projection indices, so $$f$$ cannot factor through any of the projections to finite products of $$2$$.

The argument fails more generally, though, since $$f$$ cannot be continuous for any choice of points of $$X$$ when $$X$$ is Hausdorff, say, since the subspace consisting of the points $$\{0,1\}$$ is necessarily discrete in that case, and $$f^{-1}(1)$$ is not open in $$\prod_{n \in \mathbb{N}} 2$$.

Instead, consider $$Y_0 := \{(x,y) \in \mathbb{N} \times \mathbb{N} \mid x+y \geq 0\},$$ topologized with basic open sets $$U_{a,b} := \{(x,y) \mid x \geq a, \, y \geq b\}$$. Then we have a directed collection of subspaces, $$Y_n := \{(x,y) \in Y_0 \mid 2^n \text{ divides } x+y \};$$ these subspaces are all connected, but their intersection is the subspace $$Y_{\infty} := \{(x,y) \in \mathbb{N} \times \mathbb{N} \mid x+y = 0\},$$ homeomorphic to $$\mathbb{N}$$ with the discrete topology. Suppose $$X$$ has a pair of points $$\{0,1\}$$ such that the subspace on these points is discrete. Then we have a continuous map $$Y_{\infty} \to X$$ sending $$(x,y)$$ to $$0$$ if $$y \geq x$$ and to $$1$$ otherwise. This function cannot factor through any of the $$Y_n$$, though, since the inverse images of $$0$$ and $$1$$ would have to non-trivially disconnect $$Y_n$$, which is impossible.

In summary: Since we have exhausted the possibilities for two-point subspaces, and the argument for the emptyset carries across, we have shown that in $$\mathbf{Top}$$, just as in $$\mathbf{Set}$$, the singleton space is the only strongly cocompact space.

As a final observation, if $$R$$ is a functor having a left adjoint $$L$$ which preserves cofiltered limits (if $$L$$ has a further left adjoint, for example) then $$R$$ preserves strongly cocompact objects. Indeed, letting $$X$$ be strongly cocompact and $$I$$ be a cofiltered indexing diagram, we have: $$\mathrm{Hom}(\lim_{i \in I}Y_i,R(X)) \cong \mathrm{Hom}(L(\lim_{i \in I}Y_i),X) \cong \mathrm{Hom}(\lim_{i \in I}L(Y_i),X)$$ $$\cong \mathrm{colim}_{i \in I}\mathrm{Hom}(L(Y_i),X) \cong \mathrm{colim}_{i \in I}\mathrm{Hom}(Y_i,R(X)),$$ so even without the explicit calculations in $$\mathbf{Top}$$, we could have deduced that the terminal space $$1$$ is strongly cocompact, since it is the image of the one-element set under the indiscrete functor $$\mathbf{Set} \to \mathbf{Top}$$ mentioned earlier.