Why are free Boolean topological groups Hausdorff?

Question

Assume $X$ is a Tychonoff space. Then $A(X)$ is the free topological abelian group over $X$. I know that $A(X)$ is Hausdorff and the canonical embedding from $X$ to $A(X)$ is a topological embedding.

Now consider the subgroup $N:=2A(X)=\left\{g+g\ \colon\ g\in A(X)\right\}$. The quotient $B(X):=A(X)/N$ is the free topological Boolean group. I want to show that $B(X)$ is Hausdorff which is equivalent to say that $N$ is closed in $A(X)$. Unfortunately, I do not know how show the closedness of $N$ since I can only describe the topology on $A(X)$ by the universal property and I do not know how that helps me here... Or is this maybe not true for a general Tychonoff space $X$ and one needs additional properties on $X$ like hereditarily disconnected?

I have found a reference on Free Boolean Groups

Genze, L.V. Free Boolean topological groups. Vestn. Tomsk. Gos. Univ. 2006, 290, 11–13

which maybe could answer my question but I am unable to find this text on math.sci.net (or anywhere else).

Thank you!

Answer 1

You can use the universality property with the following Boolean group as codomain: $B$ is the measure algebra over the unit interval (the quotient of the $\sigma$-algebra of Lebesgue measurable sets by the ideal of sets of measure zero), with symmetric difference as operation this is a Boolean group and $d(A,B)=\lambda(A\Delta B)$ defines a metric that will turn it into a topological group. This group contains a copy of the unit interval, namely the set $\{[0,t]:0\le t\le 1\}$. Since $X$ is Tychonoff you now have enough continuous functions to separate all non-trivial words from the empty word. This survey by Sipacheva gives more information: Free Boolean Topological Groups, Axioms 4 (2015) 492-517