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

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

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

is defined by

i ( x ) ( f ) = f ( x ) 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 ] ) [0,1]^{C(X,[0,1])} is a compact Hausdorff space. In my real analysis class I had the kids show that if X X is compact Hausdorff, the map is an embedding of X X into the cube [ 0 , 1 ] C ( X , [ 0 , 1 ] ) [0,1]^{C(X,[0,1])} . That is, the image of i i with its subspace topology is homeomorphic to X 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 → [ 0 , 1 ] C ( X , [ 0 , 1 ] ) i: X \to [0,1]^{C(X,[0,1])}

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

This leads to:

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

U : Tych → Top U: Tych \to Top

has a left adjoint

F : Top → Tych 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 Tych is a reflective subcategory of Top 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 → Y f: X\to Y factors as

X ⟶ f im ( f ) ⟶ j Y 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 → im ( f ) f: X \to im(f) . Another is the subspace topology coming from the inclusion j : im ( f ) → Y 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 Y , while the subspace topology knows nothing about the topology of X X . Indeed, suppose X X is the real line ℝ \mathbb{R} with its discrete topology, Y Y is ℝ \mathbb{R} with its codiscrete topology, and f f is the identity function. Then im ( f ) = ℝ im(f) = \mathbb{R} . With the quotient topology it’s ℝ \mathbb{R} with its discrete topology, while with the subspace topology it’s ℝ \mathbb{R} with its codiscrete topology!

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

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

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

e f(y) .

Now, a completely Hausdorff space is a topological space where any two distinct points x x and y y are separated by a continuous function f : → [ 0 , 1 ] f: \to [0,1] . So, it seems that im ( i ) 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 Top, Tych , and CompHaus CompHaus , the category of completely Hausdorff space and continuous maps.

Conjecture. There are functors

Tych ⟶ U 1 CompHaus ⟶ U 2 Top Tych \stackrel{U_1}{\longrightarrow} CompHaus \stackrel{U_2}{\longrightarrow} Top

with left adjoints

Top ⟶ F 2 CompHaus ⟶ F 1 Tych Top \stackrel{F_2}{\longrightarrow} CompHaus \stackrel{F_1}{\longrightarrow} Tych

Given any topological space X X , F 2 ( X ) F_2(X) is im ( i ) im(i) with its quotient topology, while F 1 ( F 2 ( X ) ) F_1(F_2(X)) is im ( i ) 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 : → [ 0 , 1 ] 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 → ℝ 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.