Why free topological groups on Tychonoff spaces? - MathOverflow

Why free topological groups on Tychonoff spaces?

2010-05-08T17:03:20Z

This is a question of the motivation for a common assumption found in the literature.

The free topological group $F(X)$ on a space $X$ exists for all spaces $X$ (It seems this was first shown by Katutani and Samuel). I mean "free topological group" in the sense that $F:Top\rightarrow TG$ is left adjoint to the forgetful functor $U:TG\rightarrow Top$ from the category of topological groups to the category of topological spaces.

$F(X)$ is well studied when $X$ is a Tychonoff space. This permits the application of pseudometrics which seems to be a powerful tool for describing the complicated topological structure of $F(X)$. Also, it seems to be a useful fact that the canonical map $\sigma:X\rightarrow F(X)$ is an embedding when $X$ is Tychonoff.

These two conveniences do seem to make it convenient to study $F(X)$ when $X$ is Tychonoff but it seems almost no one is interested in $F(X)$ when $X$ is not Tychonoff. Why is this? Are these uninteresting for some reason?

Answer by Johannes Hahn for Why free topological groups on Tychonoff spaces?

2010-05-08T17:44:18Z

Hi.

This is mainly because one wants to have $X$ as a subspace of $F(X)$. Since every topological group is tychonoff (:=closed sets can be seperated from points outside by continouos functions) and so is every subspace. So being Tychonoff is necessary (and sufficient) for $X$ being a subspace of $F(X)$.

Answer by Matthew Morrow for Why free topological groups on Tychonoff spaces?

2010-05-08T19:26:08Z

Let $X$ be a topological space. If $F(X)$ is $T_0$ then I think $F(X)$ is isomorphic (as topological groups) to $F(Y)$, where $Y$ is the Tychonofficiation (see below) of $X$. So it is enough to study topological free groups on a Tychonoff space.

Explanation:

First let me remind myself about some notation. Completely regular means that any point can be separated from a closed set not containing it by a continuous real-valued function. Tychonoff then means completely regular and $T_2$(=Hausdorff). Any topological group is completely regular; and for topological groups $T_0$ is equivalent to $T_2$.

Suppose $X$ is an arbitrary topological space, and let $Y$ be its "Tychonoffication" (!). That is, set theoretically $Y$ is the quotient of $X$ by the equivalence relation $x\sim x'$ if and only $f(x)=f(x')$ for all continuous $f:X\to\mathbb{R}$; each such $f$ descends to $Y$ and we give $Y$ the weak topology induced by all these real-valued maps. This makes $Y$ into a Tychonoff space which I think satisfies the following universal property: any continuous map from $X$ to a Tychonoff space factors uniquely through $Y$. Also, the natural map $X\to Y$ induces an isomorphism $C(Y)\stackrel{\cong}{\to} C(X)$, where $C(-)$ denotes the ring of real-valued continuous functions.

Assuming $F(X)$ is $T_0$, then the natural map $F(X)\to F(Y)$ seems to be an isomorphism of topological groups. It is enough to construct an inverse. Since $F(X)$ is $T_0$, it is even $T_2$, and therefore it is Tychonoff. So the natural map $X\to F(X)$ factors through $Y$ and induces $F(Y)\to F(X)$, which surely does the trick?

What about that $T_0$ assumption?

Lots of people are only interested in Hausdorff topological groups, so it seems reasonable to only study spaces $X$ for which $F(X)$ is $T_0$ (hence $T_2$). Otherwise you could replace $Y$ by the "complete-regularization" of $X$ (i.e. $X$ equipped with the weak topology induced by $C(X)$) and repeat the argument, but it doesn't work so nicely.

Edit:

While I was typing my answer, you asked about this Tychonoffication business!