February 20, 2019

Tychonoffication

Posted by John Baez Joshua Meyers is a grad student in my real analysis class. We had an interesting conversation about topology and came up with some conjectures. Maybe someone has already proved them. I just want to write them down somewhere.

First some background. For every topological space $X$ there’s a set $C(X,[0,1])$ consisting of all continuous functions from $X$ to $[0,1]$. And there’s a natural map from $X$ into $[0,1]^{C(X,[0,1])}$. The ‘cube’ $[0,1]^{C(X,[0,1])}$ is the product of copies of $[0,1]$, one for each continuous function from $X$ to $[0,1]$. So, it gets the product topology. The natural map sends each point $X$ in $X$ to the function sending each continuous function $f: X \to [0,1]$ to $f(x)$.

Get it? Maybe an equation will help. The natural map

$i: X \to [0,1]^{C(X,[0,1])}$

is defined by

$i(x)(f) = f(x)$

This is a standard ‘role reversal’ trick, turning a function into the argument and the argument into the function.

By Tychonoff’s theorem, the cube $[0,1]^{C(X,[0,1])}$ is a compact Hausdorff space. In my real analysis class I had the kids show that if $X$ is compact Hausdorff, the map is an embedding of $X$ into the cube $[0,1]^{C(X,[0,1])}$. That is, the image of $i$ with its subspace topology is homeomorphic to $X$.

So, every compact Hausdorff space is homeomorphic to a subspace of a ‘cube’: a product of copies of [0,1].

That’s the background. Then Joshua Meyers told me that more generally, a Tychonoff space can be defined as a space that’s homeomorphic to a subspace of a cube. A Tychonoff space needs to be Hausdorff (since a cube is), but it doesn’t need to be compact (since you can embed an open interval in a cube). The usual definition of Tychonoff space looks complicated and arbitrary, but it’s equivalent to this one; the definition I gave should be the definition, and the usual definition should be a theorem.

Now for the interesting part. Any space has a ‘Tychonoffication’. That’s not a very pretty word, but it’s extremely easy to guess what it means!

Namely, given any space X, we define the image of

$i: X \to [0,1]^{C(X,[0,1])}$

with its subspace topology, to be the Tychonoffication of $X$. It’s obviously a Tychonoff space, and there’s obviously a continuous map from $X$ to its Tychonoffication, namely $i: X \to im(i)$.

Conjecture. Let $Tych$ be the category of Tychonoff spaces and continuous maps. The forgetful functor

$U: Tych \to Top$

$F: Top \to Tych$

and this left adjoint is Tychonoffication. The above map from any topological space to its Tychonoffication is the unit of the adjunction. $Tych$ is a reflective subcategory of $Top$.

We went further and guessed that this adjunction factors as the composite of two other adjunctions. For this, note that any continuous map between topological space $f: X\to Y$ factors as

$X \stackrel{f}{\longrightarrow} im(f) \stackrel{j}{\longrightarrow} Y$

But we can give im(f) two different topologies. One is the quotient topology coming from the surjection $f: X \to im(f)$. Another is the subspace topology coming from the inclusion $j: im(f) \to Y$.

These topologies don’t need to be the same: after all, the quotient topology knows nothing about the topology of $Y$, while the subspace topology knows nothing about the topology of $X$. Indeed, suppose $X$ is the real line $\mathbb{R}$ with its discrete topology, $Y$ is $\mathbb{R}$ with its codiscrete topology, and $f$ is the identity function. Then $im(f) = \mathbb{R}$. With the quotient topology it’s $\mathbb{R}$ with its discrete topology, while with the subspace topology it’s $\mathbb{R}$ with its codiscrete topology!

There is always a continuous map from $im(f)$ with its quotient topology to $im(f)$ with its subspace topology. We can apply this when $f$ is the natural map

$i: X \to [0,1]^{C(X,[0,1])}$

$im(i)$ with its subspace topology is the Tychonoffication of $X$. But we can also give $im(i)$ its quotient topology. Different points $x, y \in X$ will get mapped to the same point of $im(i)$ iff these points are not separated by any continuous function $f: X \to [0,1]$, i.e. we cannot find $f$ with $f(x)\ne f(y)$.

Now, a completely Hausdorff space is a topological space where any two distinct points $x$ and $y$ are separated by a continuous function $f: \to [0,1]$. So, it seems that $im(i)$ with its quotient topology is the ‘complete Hausdorffication’ of X.

So, Joshua Meyers and I seem to believe something like this. There are three categories: $Top, Tych$, and $CompHaus$, the category of completely Hausdorff space and continuous maps.

Conjecture. There are functors

$Tych \stackrel{U_1}{\longrightarrow} CompHaus \stackrel{U_2}{\longrightarrow} Top$

$Top \stackrel{F_2}{\longrightarrow} CompHaus \stackrel{F_1}{\longrightarrow} Tych$

Given any topological space $X$, $F_2(X)$ is $im(i)$ with its quotient topology, while $F_1(F_2(X))$ is $im(i)$ with its subspace topology.

So, we’re breaking up Tychonoffication into two steps. The first step is complete Hausdorffication: it identifies points that can’t be separated by any continuous function $f: \to [0,1]$. (By the way, these are the same as the points that can’t be separated by any continous function $f : X \to \mathbb{R}$.) The second step may coarsen the topology—I don’t have a really clear mental image of what’s going on here, but a suitable example should clarify it!

I am feeling too lazy to prove these conjectures, since this has nothing to do with my main line of work. So, if anyone wants to prove them—or find a proof in the existing literature—I’d be very happy! Please let me know.

Posted at February 20, 2019 5:17 PM UTC

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Re: Tychonoffication

There’s another way to factor this adjunction (conjecturally)! We can give $X$ the weakest topology that makes $i$ continuous; let’s use $X'$ to denote $X$ with this topology. Then $X'$ is completely regular. You can define this condition by saying that every closed set and point can be separated by a continuous real-valued function (i.e. for each closed $A$ and point $p\notin A$ there is a continuous real-valued $f:X\rightarrow \mathbb{R}$ such that $f|_A$ is constantly $c\neq f(p)$).

So $i$ then factors to $X\rightarrow X'\rightarrow i(X)$. We can conjecture that the adjunction $\mathrm{Tych}\leftrightharpoons \mathrm{Top}$ then factors as $\mathrm{Tych}\leftrightharpoons \mathrm{CompReg}\leftrightharpoons \mathrm{Top}$ in addition to $\mathrm{Tych}\leftrightharpoons \mathrm{CompHaus}\leftrightharpoons \mathrm{Top}$

giving a commutative square (I would draw it but I don’t feel like trying to make $\leftrightharpoons$ vertical).

Posted by: meygerjos on February 20, 2019 7:57 PM | Permalink | Reply to this

Re: Tychonoffication

Something looks odd to me. Did you mean to write $U_1: Tych \to CompHaus$? If $U$’s are reserved for “underlying/forgetful” functors, then I would have thought $U_1: CompHaus \to Tych$, i.e., every compact Hausdorff is in fact Tychonoff.

Posted by: Todd Trimble on February 20, 2019 10:37 PM | Permalink | Reply to this

Re: Tychonoffication

CompHaus is the category of completely Hausdorff spaces, not compact Hausdorff spaces.

Posted by: meygerjos on February 20, 2019 11:19 PM | Permalink | Reply to this

Re: Tychonoffication

Posted by: Todd Trimble on February 20, 2019 11:41 PM | Permalink | Reply to this

Re: Tychonoffication

Sorry, I didn’t notice how much my category of completely Hausdorff spaces, $CompHaus$, looks like the category of compact Hausdorff spaces. I’m not really doing anything with compactness in this post, though it’s motivated by the theorem saying that every compact Hausdorff space embeds in a cube.

Posted by: John Baez on February 21, 2019 4:39 PM | Permalink | Reply to this

Re: Tychonoffication

The $F: Top \to CompHaus$ is usually called the Stone-Cech compactification, and it is the closure of the image of $X \to I^{\hom(X, I)}$ with the subspace topology. See e.g. the nLab.

Posted by: Todd Trimble on February 20, 2019 10:50 PM | Permalink | Reply to this

Re: Tychonoffication

Just so passersby can follow what’s going on: here you’re using $CompHaus$ to mean the category of compact Hausdorff spaces. Given a topological space $X$, its image under $i: X \to [0,1]^{C(X,[0,1])}$ is the universal way of making it into a Tychonoff space. So you’re saying we can go further and make in into a compact Hausdorff space in a universal way by taking the closure of the image. All this is very nice!

Usually I’ve seen Stone-Cech compactification done to Hausdorff spaces, but $X$ doesn’t need to be Hausdorff for all this stuff to work. When we take the image of $X$ under $i$, we’re forcing it to be Hausdorff—as part of making it Tychonoff.

Posted by: John Baez on February 21, 2019 4:48 PM | Permalink | Reply to this

Re: Tychonoffication

In any case, if $Y$ is Tychonoff and $f: X \to Y$, then we have a naturality square

$\array{ X & \stackrel{i_X}{\to} & I^{\hom(X, I)} \\ \mathllap{f}\; \downarrow & \, & \downarrow \mathrlap{I^{\hom(f, I)}} \\ Y & \stackrel{i_Y}{\to} & I^{\hom(Y, I)} }$

and the image of $i_X$ as a subspace of $I^{\hom(X, I)}$ can be identified with the equalizer of its cokernel pair. (The cokernel pair is formed by taking the pushout

$\array{ X & \stackrel{i_X}{\to} & I^{\hom(X, I)} \\ \downarrow \mathrlap{i_X} & \; & \downarrow \\ I^{\hom(X, I)} & \to & K_X }$

in $Top$, and regarding the unlabeled arrows in this pushout as a parallel pair of maps $I^{\hom(X, I)} \rightrightarrows K_X$.) One can easily check there is a map $K_f: K_X \to K_Y$ induced by $f$, and there is a serially commutative diagram

$\array{ I^{\hom(X, I)} & \rightrightarrows & K_X \\ \downarrow \mathrlap{I^{\hom(f, I)}} & \; & \downarrow \mathrlap{K_f} \\ I^{\hom(Y, I)} & \rightrightarrows & K_Y }$

so that by taking equalizers of the cokernel pairs, we get a commutative diagram

$\array{ im(i_X) & \to & I^{\hom(X, I)} & \rightrightarrows & K_X \\ \downarrow & \; & \downarrow \mathrlap{I^{\hom(f, I)}} & \; & \downarrow \mathrlap{K_f} \\ Y & \stackrel{i_Y}{\to} & I^{\hom(Y, I)} & \rightrightarrows & K_Y }$

where the vertical map on the left gives a factoring of $f$ through the surjective map $q_X: X \to im(i_X)$. The uniqueness of this factoring follows from surjectivity. This proves that $im(i_X)$ is indeed the Tychonoffication of $X$.

Posted by: Todd Trimble on February 20, 2019 11:36 PM | Permalink | Reply to this

Re: Tychonoffication

I mean, that $im(i_X)$ is the left adjoint of the forgetful functor $Tych \to Top$.

Posted by: Todd Trimble on February 20, 2019 11:40 PM | Permalink | Reply to this

Re: Tychonoffication

Nice, Todd!

So it sounds like you understand something I did better than I do. Namely, in this post I noticed that any continuous map

$f: X \to Y$

has a natural factorization

$X \to A \to B \to Y$

where $A$ is $im(f)$ with the topology it gets from being a quotient space of $X$, and $B$ is $im(f)$ with the topology it gets from being a subspace of $Y$.

To get $A$ from $X$ we identify points but otherwise change the topology as little as possible, while to get $B$ from $A$ we don’t change the underlying set at all, but we make the topology coarser.

I was going to ask what this is an example of. (As Kennedy proclaimed, after he became a category theorist: “Ask not for an example of this — ask what this is an example of!”)

It sounds like you know, and it sounds like the answer involves the phrase “equalizer of the cokernel pair”. What’s the general story? What sort of category has this sort of 3-step factorization for every morphism?

Posted by: John Baez on February 21, 2019 10:10 PM | Permalink | Reply to this

Re: Tychonoffication

So I think that every finitely complete, finitely cocomplete category has this sort of 3-step factorization, but often you don’t see it because often that middle step collapses to an isomorphism.

Every morphism in such a category factors as a regular epi followed by a something in the middle followed by a regular monomorphism. (Side remark: in $Top$, regular monos are the same as subspace inclusions, and regular epis are the same as quotient projections.) That is, every morphism $f: X \to Y$ factors as

$X \stackrel{p}{\to} coim(f) \to im(f) \stackrel{i}{\to} Y$

where I define $im(f)$ to be the equalizer of the cokernel pair of $f$, and $coim(f)$ to be the coequalizer of its kernel pair.

To deduce the thing in the middle, let’s give the name $q$ to the map $X \to im(f)$ which exists by the universal property of $im(f)$ as equalizer of cokernel pair plus the fact that $f$ equalizes its cokernel pair. Let’s call $\pi_1, \pi_2: E \rightrightarrows X$ the kernel pair of $f$. I claim that $q\pi_1 = q\pi_2$. That’s easy because we have

$i q\pi_1 = f\pi_1 = f\pi_2 = i q\pi_2$

and now use the fact that $i$ is monic to deduce $q\pi_1 = q\pi_2$. So then $q: X \to im(f)$ must factor through the coequalizer of the kernel pair $(\pi_1, \pi_2)$, which is $p: X \to coim(f)$; that factoring gives the thing in the middle $coim(f) \to im(f)$.

Anyway, applying these general observations to the unit $i_X: X \to I^{\hom(X, I)}$ of the monad you introduced, the coimage of $i_X$ is this quotient topology on $X$ that you’re proposing as the complete Hausdorffication (and I think you’re right about that), and the image $i_X$ is of course the Tychonoffication. But I am also supposing that there is a much more general nonsense story that might be told, abstracting away from the particular topological context we are discussing. I don’t know what the story is, exactly, yet.

By the way, I’m not sure which definition of Tychonoff you meant that looks arbitrary and scary, but the nLab defines a Tychonoff space to be any space arising as a subspace of a compact Hausdorff space, and I think that’s actually a very nice flexible definition. Shot through the story here is the fact that $I$ is a cogenerator in the category of compact Hausdorff spaces, something closely connected with the Urysohn lemma and such.

Posted by: Todd Trimble on February 21, 2019 11:53 PM | Permalink | Reply to this

Re: Tychonoffication

The “arbitrary and scary” definition of Tychonoff space is the one used on Wikipedia: a Tychonoff space is a completely regular Hausdorff space. Here a completely regular topological space is one where points can be separated by closed sets by continuous real-valued functions.

This is not actually very scary, but “a subspace of a product of closed intervals” is better and “a subspace of a compact Hausdorff space” is even better.

It’s actually great to have both the extrinsic definition (characterizing it as a subspace of something) and the intrinsic one.

Posted by: John Baez on February 22, 2019 6:05 AM | Permalink | Reply to this

Re: Tychonoffication

My favorite definition of Tychonoff space is that for any open set $U\in \mathcal{O}(X)$ we have $U = \bigcup \{ V\in \mathcal{O}(X) \mid V \lt\!\!\!\!\lt\!\!\!\!\lt U \}$, where $V\lt\!\!\!\!\lt\!\!\!\!\lt U$ means that the closure of $V$ is contained in $U$ and that there exists $W\in \mathcal{O}(X)$ such that $V\lt\!\!\!\!\lt\!\!\!\!\lt W \lt\!\!\!\!\lt\!\!\!\!\lt U$ (coinductively).

Posted by: Mike Shulman on February 22, 2019 7:39 AM | Permalink | Reply to this

Re: Tychonoffication

Admittedly, that coalgebraic perspective is pretty sweet. I never thought of approaching the Urysohn lemma from that point of view.

Posted by: Todd Trimble on February 23, 2019 2:53 AM | Permalink | Reply to this

Re: Tychonoffication

Posted by: Mike Shulman on February 22, 2019 12:20 AM | Permalink | Reply to this

Re: Tychonoffication

That’s nice! You probably remember us writing about the weak ternary factorization system in $Cat$ in Lectures on n-categories and cohomology. Given a functor $p: E \to B$ we factored it thus:

We start with $E$; then we throw in new 2-morphisms (equations between morphisms) that we get from $B$; then we throw in new 1-morphisms (morphisms), and finally new 0-morphisms (objects). It’s like a horse transforming into a person from the head down. First it’s a horse, then it’s a centaur, then it’s a faun-like thing that’s horse from the legs down, and finally it’s a person.

But I’d never thought about the strict ternary factorization system in $Top$!

Posted by: John Baez on February 22, 2019 6:12 AM | Permalink | Reply to this

Re: Tychonoffication

This sounds like something that ought to be in Johnstone’s book Stone Spaces, but I don’t have it with me at the moment to look.

It also sounds like we ought to be saying something about codensity monads.

Posted by: Mike Shulman on February 21, 2019 6:03 PM | Permalink | Reply to this

Re: Tychonoffication

Apologies in advance for pedantry verging on sealioning, but shouldn’t it be “Tychonoffification”, just as it should be “Hausdorffification”?

(Corrections/confirmation from e.g. Blake Stacey welcome.)

Posted by: Yemon Choi on February 21, 2019 8:19 PM | Permalink | Reply to this

Re: Tychonoffication

That was exactly my first thought on seeing this post.

(And thanks, Yemon, for introducing me to the term sealioning. I remember once reading the comic that inspired the term, but hadn’t encountered the word.)

Posted by: Mark Meckes on February 21, 2019 8:56 PM | Permalink | Reply to this

Re: Tychonoffication

I think it should really be “Tychonoffifification”, just to make it easier to say.

Posted by: John Baez on February 21, 2019 9:59 PM | Permalink | Reply to this

Re: Tychonoffication

By the way, Tychonoff spaces are sometimes called $T_ {3\frac{1}{2}}$ spaces — or even $T_\pi$ spaces! That’s cute, though admittedly not very rational.

Posted by: John Baez on February 22, 2019 6:47 AM | Permalink | Reply to this

Re: Tychonoffication

I see what you did there.

More seriously, despite the inherent arbitrariness of the scale, when I was first learning about the various separation axioms, I found the $T_n$ terminology easier to keep straight than “Hausdorff”, “regular”, “normal”, etc. How are you supposed to remember whether “regular” or “normal” is meant to be nicer, let alone how “Hausdorff” fits in there?

Posted by: Mark Meckes on February 22, 2019 11:50 AM | Permalink | Reply to this

Re: Tychonoffication

I’ve never really managed to remember the $T_n$ terminology, though I probably knew it for a few months in my youth. It’s just too unevocative for me.

I know a bunch of grad students who fall in love with point set topology and separation axioms and have to be “talked down”, like: “yes, this is fun stuff, but it’s not fashionable, since it’s hard to do anything new in this area that seriously affects other subjects…”

I don’t recall going through that phase myself.

Posted by: John Baez on February 22, 2019 5:20 PM | Permalink | Reply to this

Re: Tychonoffication

I don’t claim to be able to remember the $T_n$ terminology now. But “regular” and “normal” aren’t evocative enough to tell me which is which either. It is at least easier to remember that $T_4$ is stronger than $T_3$ than it is to remember that “normal” is stronger than “regular”.

Fortunately, mostly I’ve never needed to remember any of this stuff — at least not since I took a qualifying exam on which I had to prove that a topological space is Tychonoff if and only if it’s homeomorphic to a subspace of a cube.

Posted by: Mark Meckes on February 22, 2019 6:58 PM | Permalink | Reply to this

Re: Tychonoffication

I didn’t get sucked into this stuff myself as a graduate students; I think it was only with nLab work did I ever take a closer look.

For me, the main virtue of the $T_n$ terminology is that it serves to remind me that they all include the $T_0$ condition, that points are distinguished by the systems of neighborhoods containing them. A uniform structure plus $T_0$ gets you all the way to $T_{3\frac1{2}}$ (if I remember correctly), and pretty soon after that point I begin to lose interest, although the category of $T_4$ spaces (which are normal $T_2$ spaces) has IIRC some pretty decent categorical properties.

I think I learned a lot of this stuff indirectly through Toby Bartels. He wrote a lot of the nLab material on separation axioms.

From a career perspective John is probably right to talk down those students, but quite a lot of point-set topology is just so weird and fascinating and charming; it’s easy to understand why those grad students can get sucked in.

Posted by: Todd Trimble on February 23, 2019 2:50 AM | Permalink | Reply to this

Re: Tychonoffication

This is a very interesting post for quantum gravity theorists. We are in the business of deriving the emergent (classical) manifolds from quantum information, using categorical operads like the n-cubes operad in each dimension. Quantum logic is infinite dimensional because the cardinality of a set is replaced by the dimension of a vector space, as a rough idea. Amazing that Hilbert somehow had this deep intuition so long ago, but then much about quantum mechanics was swept under the rug in the twentieth century!

Posted by: Marni Dee Sheppeard on February 22, 2019 10:54 PM | Permalink | Reply to this

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