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.
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.