# Why free topological groups on Tychonoff spaces?

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?

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–  Martin Brandenburg May 8 '10 at 18:33

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:

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Thanks Matthew. Might anyone suggest a reference for "Tychonoffication?" –  Jeremy Brazas May 10 '10 at 2:20
The bible for this sort of stuff: Gillman & Jerison, "Rings of Continuous Functions". Chapter 1. Hopefully it is in your library; otherwise it is a hard book to get your hands on. –  Matthew Morrow May 10 '10 at 12:25
I am actually a bit hesitant about this now...Theorem 1 of B.V.S Thomas' paper "Free Topological Groups" in Lecture Notes in Mathematics Volume 378/1974 says that $F(X)$ is Hausdorff if and only if $X$ is functionally Hausdorff (sometimes known as completely Hausdorff). This answers my follow up question to Johannes. It seems to me that the "complete-regularization" of a functionally Hausdorff space night not always give you back $X$ as the topology might not be the weak topology from maps $X\rightarrow \mathbb{R}$ (i.e. not all functionally Hausdorff spaces are completely regular). –  Jeremy Brazas May 10 '10 at 13:13
This is the problem I was seeing which caused me to say "it doesn't work so nicely" in my answer (in the case when $F(X)$ is not $T_0$, which I now know is the same as $X$ being functionally Hausdorff). But I'm not sure what you are asking now. Taking into account this result of Thomas (which I didn't know), my answer says that if $X$ is functionally Hausdorff then $F(X)=F(Y)$, where $Y$ is $X$ equipped with the weak $C(X)$ topology ($Y$ is also the Tychonoffication of $X$). –  Matthew Morrow May 10 '10 at 14:30
You are right...I finally checked the details to convince myself. The "Tychonoffization" quotient $X\rightarrow Y$ does induce an isomorphism on free topological groups. Sorry about the confusion and thanks again! –  Jeremy Brazas May 10 '10 at 14:38

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 $F(X)$.

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Thanks for the answer! Is it necessarily true that $X$ is Tychonoff if and only if $F(X)$ is T0 (and necessarily Tychonoff)? Or could some non-Tychonoff space $X$ yield an interesting, Tychonoff free topological group that contains (this may be silly or ignorant... I'm not sure) a "Tychonoff-ized" copy of $X$? –  Jeremy Brazas May 8 '10 at 18:52
I didn't mention T0 and it does not matter for the facts I listed. It is all true without T0 (some call this tychonoff-minus-T0-property $T_{3 \frac{1}{2}}$). –  Johannes Hahn May 9 '10 at 1:12
We seem to be using different terminology.... I think that $T_{3\frac{1}{2}}$ means Tychonoff, which includes $T_0$. And Tychonoff minus $T_0$ means completely regular. Every topological group is completely regular, but not necessarily Tychonoff (just take a nonzero group with the trivial topology). As Johannes says, $X$ will embed into $F(X)$ if and only if $X$ is completely regular (in my terminology). Otherwise I think $F(X)$ contains a copy of the "complete-regularization" of $X$, as I mention in my answer. –  Matthew Morrow May 9 '10 at 14:11