# 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?

• May 8, 2010 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:

• Thanks Matthew. Might anyone suggest a reference for "Tychonoffication?" May 10, 2010 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. May 10, 2010 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). May 10, 2010 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 $X$). May 10, 2010 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! May 10, 2010 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 $F(X)$.

• 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 $X$? May 8, 2010 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 $T_{3 \frac{1}{2}}$). May 9, 2010 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. May 9, 2010 at 14:11