### 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 $i: X \to im(i)$.

This leads to:

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

$U: Tych \to Top$

has a left adjoint

$F: Top \to Tych$

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

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

with left adjoints

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

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.

## 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).