# Group topology and free topological groups

The following result is in Topological Groups and Related Structures, Arhangel'skii, Tkachenko;

Notation: $F_a(X)$ denotes the free group on $X$ and $F(X)$ the free topological group on $X$.

Claim: Let $f:X\to Y$ be a contiuous mapping of Tychonoff spaces. Then $f$ admits an extension to continuous homomorphism $F(f):F(X)\to F(Y)$. In addition, if $f$ is a quotient, then $F(f)$ is open.

We recall we already proved that $F(X)$ and $F_a(X)$ are the same of algebraic point of view.

Proof: Since $Y$ is identified with the corresponding subspace of $F(Y)$, we can consider $f$ as a continuous mapping of $X$ to the free topological group $F(Y)$. Therefore, by the definition of $F(x)$, $f$ can be extended to a continuous homomorphism $F(f):F(X)\to F(Y)$ wich we shall denote by $\overline{f}$.

Now suposse that $f$ is a quotient. Denote $\tau _q$ the family of all $\overline{f}(U)$ where $U$ is open in $F(X)$. Then $\tau _q$ is a group topology on the abstract group $F_a(Y)$.

Here are my questions:

1.- Why is $\tau _q$ a topology? I don't see why is closed under finite intersections.

2.- Why is $(F_a(Y),\tau _q)$ a topological group? I don't know how to prove that the map $(x,y)\mapsto xy$ is continuous on $F_a(Y)\times F_a(Y)$.

I recall that $f$ is a quotient map if it is surjective, continuous and $f^{-1}(V)$ is open implies $V$ is open.

Thanks.

• Are you the same person as math.stackexchange.com/users/73564/user73564? – user17762 May 27 '13 at 4:28
• Yes... I really don't know why I have two different accounts. I think it is because I have two PC's... anyway. By the way, I posted about this result in general topological spaces but it was false... It must be the structures of free groups were necessary or something like that. – user73564 May 27 '13 at 4:34

It seems that $(F_a(Y),\tau_q)$ is a quotient group (see section 1.5 of “Topological Groups and Related Structures” about this construction) of the topological group $F(X)$, and $\bar f$ is the quotient map.
• The group $(F_a(Y),\tau_q)$ should be a quotient group of the topological group $F(X)$ by the definition of the topology $\tau _q$ as the family of all $\overline{f}(U)$ where $U$ is open in $F(X)$. – Alex Ravsky Jun 7 '13 at 4:01