Add category of compact Hausdorff spaces

[dschepler]

Addresses: #156

Currently undecided properties:
has cartesian filtered colimits
is coaccessible
has cofiltered-limit-stable epimorphisms
is coregular
is countably distributive
has exact cofiltered limits
is infinitary distributive
is locally copresentable
has a natural numbers object

[dschepler]

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.

[ScriptRaccoon]

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.

[ScriptRaccoon]

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

[dschepler]

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 $[n] \to \beta(N)$.

[dschepler]

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.

[dschepler]

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.

[dschepler]

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.

[ScriptRaccoon]

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.

[dschepler]

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.

[dschepler]

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?

[ScriptRaccoon]

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.

[ScriptRaccoon]

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 CompHaus^op is monadic over Set. So I assume the question is if this monad is accessible.

[ScriptRaccoon]

Apparently, Isbell proved that CompHaus^op 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 CompHaus^op 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 CompHaus^op.

[ScriptRaccoon]

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]): CompHaus^op →
  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 CompHaus^op is given in [Isb82], and
• a complete description of the algebraic theory of CompHaus^op is obtained in [MR17].

[dschepler]

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?

[dschepler]

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.)

[ScriptRaccoon]

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.