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Add category of compact Hausdorff spaces#160

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May 14, 2026
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Add category of compact Hausdorff spaces#160
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@ScriptRaccoon ScriptRaccoon linked an issue May 5, 2026 that may be closed by this pull request
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Hmm, for the coregular property, I have two possible approaches:
Try to adapt the proof from Top.
Otherwise, a Google search reminded me of Gelfand duality, that the opposite category of CompHaus is equivalent to the category of commutative unital $C^$-algebras; and the nLab page on that category says the category of not necessarily unital or commutative $C^$-algebras is monadic over Set, and therefore regular. Not sure whether there's a way to get from there to regularity of the subcategory.

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ScriptRaccoon commented May 5, 2026

Thanks for the PR!

For the non-existence of NNO, and hence showing that CompHaus is not countably distributive, I suggest to use the lemma nno_distributive_criterion that I added a few hours ago (to show that SemiGrp has no NNO).

If an NNO exists, it has to be $\beta(N)$, and for every compact Hausdorff space $A$, the canonical morphism

$$\alpha : \beta(A \times N) \to A \times \beta(N)$$

needs to be a split monomorphism. But it is clearly injective and has dense image. So it would actually be an isomorphism. I don't think that this is true when $A$ is not discrete. But I have no proof, yet.

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I think coregularity should be easy to prove directly. Don't go via C*-algebras here.

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The question you raise regarding a consequence of NNO existing also happens to be a special case of whether the category has cartesian filtered colimits, for the special case of the colimit $[n] \to \beta(N)$.

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Regarding the property of having cofiltered-limit-stable epimorphisms: It's interesting that the counterexample from Set and Haus fails in CompHaus. In fact, by the intersection theorem, any such example where it's asking about an intersection of a codirected family of nonempty compact subobjects to the constant 1 is doomed to failure. I don't have any ideas on the general case, though.

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How about this: if $I = [0, 1]$, then $I^I$ does exist as an exponential in Top with the compact-open topology, it's just not compact. So, we have $Hom(\beta(N\times I), I) \simeq Hom(N\times I, I) \simeq Hom(N, I^I) \not\simeq Hom(\beta(N), I^I) \simeq Hom(\beta(N)\times I, I)$, and hopefully by analyzing the chain and a failure example at the one point, we could come up with a proof that $\beta(N\times I) \not\simeq \beta(N)\times I$. I haven't managed to get through the details yet, however.

Maybe constructing the counterexample of a sequence in $I^I$ with no convergent subnet via recursion of multiplying by $x$ would work. Again, not quite getting through the details on that yet.

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I think I've got it now: if a natural numbers object $N$ existed, we could iterate the initial conditions $I \to I\times I$, $x \mapsto (x, x)$ and $I\times I \to I \times I$, $(x, y) \mapsto (x, xy)$ to get a continuous function $N \times I \to I \times I$ such that $(n, x) \mapsto (x, x^n)$ for $n \in \mathbb{N}$. The sequence $(n) \in N$ has a convergent subnet $(n_\lambda){\lambda \in \Lambda}$, say with limit $y$, and for any $x\in I$, $(n\lambda, x) \mapsto (x, x^{n_\lambda})$. So taking limits, $(y, x) \mapsto (x, 0)$ if $x \ne 1$ or to $(x, 1)$ if $x=1$. That contradicts that the composition $x \mapsto (y, x) \mapsto (x, \delta_{x,1}) \mapsto \delta_{x,1}$ would have to be continuous.

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ScriptRaccoon commented May 6, 2026

Epimorphisms might actually be stable under cofiltered limits. This is equivalent to: monomorphisms of commutative unital C*-algebras ( = injective *-homomorphisms) are stable under filtered colimits – which looks correct, even for all C*-algebras.

More generally, exact cofiltered limits are likely. Also, locally copresentable looks promising.

But let's try to find a purely topological proof.

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dschepler commented May 6, 2026

I've found a proof of cofiltered-limit-stable epis. A brief outline: First, a lemma that any cofiltered limit of nonempty compact Hausdorff spaces is nonempty. To see this, consider the product, and for $f : i \to j$ in the cofiltered category, let $F_f := \{ x \in \prod X_i \mid X_f(x_i) = x_j \}$. Each one is closed, and it's easy to see the family has the finite intersection property (in writing it up, I'd of course expand this point a bit more). Therefore, the intersection of all $F_f$ is nonempty; but that intersection is precisely the limit.

For the main statement: just apply the lemma to the cofiltered limit of inverse images of ${ y_i }$.

I'm still not sure whether this is just a special case of a more general argument in disguise, where the more general argument establishes exact cofiltered limits.

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I think I'm getting closer to a proof that it's $\aleph_1$-coaccessible, but I'm still fuzzy on the details.

The basic idea: first prove that $[0,1]$ is $\aleph_1$-copresentable. For that, if we have $\lim X_i \to [0,1]$, then for each point of $\lim X_i$, use continuity and filtering to find an open subset of some $X_i$ whose image is within an interval of diameter at most $1/n$. Choosing a finite subcover of the limit, we can assume that all $i$ are the same. Then if we do this for each $n$, then use filtering, that should find an $i$ such that the map factors through $X_i$.

It should then follow that all countable powers of $[0,1]$ are $\aleph_1$-copresentable. And for any compact $X$, it might then be a limit of all possible maps to such countable powers?

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ScriptRaccoon commented May 7, 2026

I found an interesting paper: https://arxiv.org/pdf/1808.09738
I haven't read it. But it proves an interesting characterization of the category of compact Hausdorff spaces. Also, it gives references for the more well-known properties of this category. Maybe we can exchange the nLab references with them. And maybe this paper also contains proofs for the properties that are currently open.

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ScriptRaccoon commented May 7, 2026

https://doi.org/10.1016/j.topol.2019.02.033 proves several results about locally copresentable categories of spaces (called: dually locally presentable categories). See Theorem 3.4 and Theorem 4.9. It appears that CompHaus is never mentioned explicitly, but I assume this is because the authors are way past that example :D. I will find a better reference.

Close catch: https://arxiv.org/pdf/1508.07750 - references in particular the result that CompHausop is monadic over Set. So I assume the question is if this monad is accessible.

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ScriptRaccoon commented May 7, 2026

Apparently, Isbell proved that CompHausop is equivalent to the category of functors $T \to Set$ preserving countable products, where $T$ is the ess. small subcategory of CompHaus consisting of all countable powers of the unit interval. So CompHausop is actually countable-ary algebraic. From here, it should be clear that the category is locally $\aleph_1$-presentable.

J. Isbell. Generating the algebraic theory of C(X). Algebra Universalis, 15(2):153–155, 1982

I don't have access to the paper. And tbh the papers of Isbell are never easy to understand. So I would appreciate if we can find a more direct argument for local $\aleph_1$-presentability of CompHausop.

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ScriptRaccoon commented May 7, 2026

The introduction of http://www.tac.mta.ca/tac/volumes/33/12/33-12.pdf gives useful references.

  • in [Dus69] it is proved that the representable functor hom(−, [0, 1]): CompHausop
    Set is monadic,
  • the unit interval [0, 1] is shown to be a ℵ1-copresentable compact Hausdorff space
    in [GU71],
  • a presentation of the algebra operations of CompHausop is given in [Isb82], and
  • a complete description of the algebraic theory of CompHausop is obtained in [MR17].

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To prove a category has exact cofiltered limits, is it sufficient to show that cofiltered limits commute with binary coproducts, initial objects, and coequalizers? Or is there something subtle there that I'm missing?

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dschepler commented May 8, 2026

I think I have a proof that the functor Hom(-, [0,1]) : CompHaus^op -> Set is monadic, using the crude monadicity criterion: suppose we have a coreflexive equalizer $E \to A \rightrightarrows B$ with $r : B \to A$. My mental picture of this uses $r$ to view $B$ as a bundle of compact spaces over $A$, and $f, g : A \rightrightarrows B$ as two sections of that bundle. Then in $Hom(B,[0,1]) \rightrightarrows Hom(A,[0,1]) \to Hom(E,[0,1])$, define $s : Hom(E,[0,1])\to Hom(A,[0,1])$ as a choice function of Tietze extensions. Now given $h : A \to [0,1]$, we can define a continuous function on $im(f)\cup im(g) \simeq A +_E A$ to be $h$ on the first $A$, and to be $s(h|_E)$ on the second $A$. Then again choosing a Tietze extension of this to $B$ for each $h$ gives a function $t : Hom(B,[0,1]) \to Hom(A,[0,1])$. By construction, $s,t$ make $Hom(E,[0,1])$ a split coequalizer.

As for showing it is conservative: suppose $f : X \to Y$ such that $Hom(Y,[0,1]) \to Hom(X,[0,1])$ is a bijection. Then for $x_1\ne x_2$ in $X$, choose a function $\varphi : X \to [0,1]$ with $\varphi(x_1) = 0$, $\varphi(x_2) = 1$. Then since there is $\psi : Y \to [0,1]$ such that $\psi \circ f = \varphi$, we get $f(x_1) \ne f(x_2)$. Therefore, $f$ is injective. Similarly to the proof of epimorphisms being surjective, we can also see $f$ is surjective. So, since $f$ is a bijective continuous map between compact Hausdorff spaces, it is a homeomorphism.

Oh, and for having a left adjoint: that's the functor $S \mapsto [0,1]^S$.

Combined with $\aleph_1$-accessibility of $[0,1]$, that should prove CompHaus^op is locally $\aleph_1$-presentable. (Though I just realized I was missing half of the proof, that if you have two morphisms $X_i \to Y$ which become equal in the limit, then they're already equalized by some $X_f, X_g : X_j \to X_i$. That should be doable by considering sets where the two functions differ by at least $1/n$; use the previous lemma on filtered colimits of nonempty spaces, in contrapositive, to show every point there is not in the image of some $X_f$; use compactness to show the whole part where they differ by at least $1/n$ is not in the image of some $X_f$; then use countable filtering to conclude there's some $X_f$ whose image is entirely in the equalizer.)

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ScriptRaccoon commented May 8, 2026

To prove a category has exact cofiltered limits, is it sufficient to show that cofiltered limits commute with binary coproducts, initial objects, and coequalizers? Or is there something subtle there that I'm missing?

Yes this works as soon as cofiltered limits and finite colimits exist, otherwise they cannot be computed pointwise e.g. and this is also why it was an assumption for the property in the first place. Also, initial objects are never a problem, as the constant diagram with value X on a non-empty category has limit X.

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ScriptRaccoon commented May 9, 2026

I would like to hear your feedback regarding the current state of #164 (the biggest PR I have worked on so far). For this I have converted the definition of CompHaus and a selection of its properties from this PR into a yaml file CompHaus.yaml, which you see below. I would very much appreciate to know what you think about this format. Is it better than the current model? What is your first impression?

id: CompHaus
name: category of compact Hausdorff spaces
notation: $\CompHaus$
objects: compact Hausdorff spaces
morphisms: continuous functions
description: |
  This is the full subcategory of $\Top$ consisting of those spaces that are [compact](https://en.wikipedia.org/wiki/Compact_space) and [Hausdorff](https://en.wikipedia.org/wiki/Hausdorff_space).
nlab_link: https://ncatlab.org/nlab/show/empty+category
tags:
  - topology

related_categories:
  - Top
  - Haus

satisfied_properties:
  - property_id: locally small
    reason: This is trivial.

  - property_id: generator
    reason: The one-point space is a generator because it represents the forgetful functor to $\Set$, which is faithful.

  - property_id: products
    reason: |
      By the Tychonoff product theorem, a product in $\Top$ of compact Hausdorff spaces is compact; it is also clearly Hausdorff.
      Since the forgetful functor from $\CompHaus$ to $\Top$ is fully faithful, this limit is reflected in $\CompHaus$ as well.

  - property_id: equalizers
    reason: |
      The equalizer in $\Top$ of two continuous functions $f, g : X \rightrightarrows Y$ between compact Hausdorff spaces is a closed subspace of $X$, and therefore it is also compact Hausdorff. Since the forgetful functor from $\CompHaus$ to $\Top$ is fully faithful, this limit is reflected in $\CompHaus$ as well.

  - property_id: cocomplete
    reason: |
      $\CompHaus$ is a reflective subcategory of $\Top$, with the reflector being the Stone-Čech compactification functor.
      See [nLab](https://ncatlab.org/nlab/show/compact+Hausdorff+space#StoneCechCompactification) for example.
      Therefore, as usual, we can form colimits in $\CompHaus$ by forming colimits in $\Top$ and then applying Stone-Čech compactification.

unsatisfied_properties:
  - property_id: natural numbers object
    reason: |
      Let $I := [0, 1]$. If a natural numbers object $(N, z : 1 \to N, s : N \to N)$ existed, then we could iterate the initial conditions $I\to I\times I$, $x \mapsto (x, x)$ and the recursive step function $I\times I \to I \times I$, $(x, y) \mapsto (x, xy)$ to get a continuous function $N \times I \to I \times I$ such that $(s^n(z), x) \mapsto (x, x^n)$ for $x\in I$, $n \in \IN$.
      The sequence $(s^n(z)) \in N$ has a convergent subnet $(s^{n_\lambda}(z))_{\lambda \in \Lambda}$, say with limit $y$. Thus, for any $x\in I$ and $\lambda \in \Lambda$, we have $(s^{n_\lambda}(z), x) \mapsto (x, x^{n_\lambda})$.
      Taking limits, we see $(y, x) \mapsto (x, 0)$ if $x \ne 1$ or $(y, x) \mapsto (x, 1)$ if $x = 1$. In other words, $(y, x) \mapsto (x, \delta_{x, 1})$ for all $x\in I$. However, that contradicts the fact that the composition
      $$I \overset{y \times \id}\longrightarrow N\times I \to I\times I \overset{p_2}\longrightarrow I \\ x \mapsto (y, x) \mapsto (x, \delta_{x,1}) \mapsto \delta_{x,1},$$
      would have to be continuous.

special_objects:
  terminal object:
    description: singleton space
  initial object:
    description: empty space

special_morphisms:
  epimorphisms:
    description: surjective continuous maps (which are automatically quotient maps)
    reason: For the non-trivial direction, and for a proof of the parenthetical remark, see the proof above that $\CompHaus$ is epi-regular.

@dschepler
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I would like to hear your feedback regarding the current state of #164 (the biggest PR I have worked on so far). For this I have converted the definition of CompHaus and a selection of its properties from this PR into a yaml file CompHaus.yaml, which you see below. I would very much appreciate to know what you think about this format. Is it better than the current model? What is your first impression?

The sample looks great, and I agree with the comments by others that having everything directly related to the category in one file should be easier to work with.

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Thank you for the feedback!

The mentioned PR is a lot of work and I wanted to make sure that I am going into the right direction. Apparently, yes.

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ScriptRaccoon commented May 10, 2026

... but I have decided against using markdown (long story...), so instead of
[Hausdorff](https://en.wikipedia.org/wiki/Hausdorff_space)
it will be
<a href="https://en.wikipedia.org/wiki/Hausdorff_space" target="_blank">Hausdorff</a>
as before. So the end result will be a bit different from the file I shared above.

@dschepler dschepler marked this pull request as ready for review May 10, 2026 19:34
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Interesting search...
image
(And the dual search returns no results.)

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I think I am finished with #164. Should I wait with the merge? As before, I am not sure which order is best. Maybe you have a preference?

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I think I am finished with #164. Should I wait with the merge? As before, I am not sure which order is best. Maybe you have a preference?

It looks like I have a significant amount of revisions to the pdf file (and a few updates otherwise) to make. So, it's probably best if you merge #164 and I can take care of updating my changes to the new YAML schema, rather than my revisions delaying the other merge for maybe a day or two.

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ScriptRaccoon commented May 11, 2026

Alright. I have merged the PR now. If you need any help with transforming the data to a YAML file, or if something is a bit weird, please let me know. The documentation has been updated as well.

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The spacing on the display of the "natural numbers object" proof is weird now. I could maybe alleviate it a bit by putting them in a single $$ line 1 \ line 2 $$ display; but then the "pnpm dev" console would constantly be displaying warnings that \ within $$ is not supported by LaTeX. And that would still result in an extra blank line between the display and the next line.

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ScriptRaccoon commented May 12, 2026

The spacing on the display of the "natural numbers object" proof is weird now. I could maybe alleviate it a bit by putting them in a single $$ line 1 \ line 2 $$ display; but then the "pnpm dev" console would constantly be displaying warnings that \ within $$ is not supported by LaTeX. And that would still result in an extra blank line between the display and the next line.

How I would do it:

  - property_id: natural numbers object
    reason: >-
      Let $I := [0, 1]$. If a natural numbers object $(N, z : 1 \to N, s : N \to N)$ existed, then we could iterate the initial conditions $I\to I\times I$, $x \mapsto (x, x)$ and the recursive step function $I\times I \to I \times I$, $(x, y) \mapsto (x, xy)$ to get a continuous function $N \times I \to I \times I$ such that $(s^n(z), x) \mapsto (x, x^n)$ for $x\in I$, $n \in \IN$. The sequence $(s^n(z)) \in N$ has a convergent subnet $(s^{n_\lambda}(z))_{\lambda \in \Lambda}$, say with limit $y$. Thus, for any $x\in I$ and $\lambda \in \Lambda$, we have $(s^{n_\lambda}(z), x) \mapsto (x, x^{n_\lambda})$. Taking limits, we see $(y, x) \mapsto (x, 0)$ if $x \ne 1$ or $(y, x) \mapsto (x, 1)$ if $x = 1$. In other words, $(y, x) \mapsto (x, \delta_{x, 1})$ for all $x\in I$. However, that contradicts the fact that the composition
      $$\begin{align*} I & \xrightarrow{y \times \id}  N \times I \xrightarrow{} I\times I \xrightarrow{p_2} I, \\ x & \mapsto (y, x) \mapsto (x, \delta_{x,1}) \mapsto \delta_{x,1}, \end{align*}$$
      would have to be continuous.

This looks fine now.

NNO proof

You can also keep the line break in the alignment. Just use >- as the marker

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I've just made more revisions to the PDF to try to clarify the arguments. Let me know if I still need more clarification.

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The merge conflict comes from #173, sorry for that.

I will have another look at this PR now.

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This shows that $\CompHaus^{\op}$ is equivalent to the category of algebras over the monad $S \mapsto \Hom_{\CompHaus}([0, 1]^S, [0, 1])$. We may view such algebras as being models of the infinitary algebraic theory of all continuous functions $[0,1]^S \to [0,1]$. In fact, we can show that any such function only depends on countably many coordinates in the domain, so that operations of this theory will be generated by the continuous functions $[0,1]^\omega \to [0,1]$. Indeed, we get the following somewhat stronger result:

\begin{proposition}
The object $[0,1]$ of $\CompHaus$ is $\aleph_1$-copresentable. \cite{GU71}
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Suggested change
The object $[0,1]$ of $\CompHaus$ is $\aleph_1$-copresentable. \cite{GU71}
The object $[0,1]$ of $\CompHaus$ is $\aleph_1$-copresentable.

... because the citation looks like this result is proven there, but I assume it's not

Or, did you want to cite a different article?

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The introduction of the paper I got the GU71 citation from said that result was proved there.

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Ah, I didn't know that this book goes into topological details. So I checked all results for compact Hausdorff spaces. (In that book, "Hausdorff" is included in "compact".)

  • 4.7 proves that $\{I \times I\}$ is dense in $CompHaus^{op}$ (which was first shown by Isbell)
  • 4.8 proves that $CompHaus$ has no small dense subcategory.
  • 6.5 (a) proves that in $CompHaus$ only the empty space is generated (in the sense of $\kappa$-generated for some cardinal $\kappa$).
  • 6.5. (b) proves that in $CompHaus^{op}$, a space is finitely generated <=> finitely presented <=> finite.
  • 6.5 (c) proves that $I$ is $\aleph_1$-presentable in $CompHaus^{op}$.
  • 9.4 (d) proves that in $CompHaus^{op}$, a space is $\aleph_1$-generated <=> $\aleph_1$-presentable <=> metrizable. (The proof reduces to the fact for $I$.)

I can translate the proof of 6.5 (c) and post it here, if you like. It is quite different. In any case, let's add the more precise reference.

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I guess I ended up reproving 6.5(a) in the proof that CompHaus is not accessible (which is no longer needed now that we know it's locally copresentable).

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Going back and rereading the introduction to one of the cited papers prompted me to refine the last paragraph of the corollary proof a bit:

We thus reproduce a result from \cite{Isb82} which also provides a nice description of a small set of generators of the operations of the $\aleph_0$-ary algebraic theory. A more recent treatment in \cite{Marra_2017} refines this by providing a nice axiomatization of the relations of that theory.

@ScriptRaccoon ScriptRaccoon merged commit ce583ae into ScriptRaccoon:main May 14, 2026
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Good work :)

What about the category of locally compact Hausdorff spaces? :D

@dschepler dschepler deleted the comphaus branch May 14, 2026 13:31
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Remark: after merging of #184 the pdf from this PR has now been removed and replaced with the markdown-rendered page https://catdat.app/content/comphaus_copresentable

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Add category of compact Hausdorff spaces

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