Homomorphism Theorem

Question

Let $f$ be an open homomorphism from a topological group $G$ onto a topological group $H.$ We denote $K=Ker(f).$
How can I prove that $\bar f:G/K→H$ is a homeomorphism? I tried to prove it is continuous but failed. Please suggest me.

Answer 1

This is because of the universal property of quotient spaces.

If $f:G\to H$ is a continuous surjection between spaces, $R$ is an equivalence relation on $G$, and $f$ respects the relation, i.e. $f(g)=f(g')$ whenever $gRg'$, then we have a unique continuous surjection $\tilde f: G/R\to H$ such that $\tilde f\circ q=f$ where $q:G\to G/R$ is the canonical surjection. If $f(g)=f(g')$ if and only if $gRg'$, then $\tilde f$ is a bijection. And if $f$ is open or closed (or more generally a quotient map), then $\tilde f$ is a homeomorphism.

In the case of groups $G,H$ and a normal subgroup $N\le G$, we have a relation $R$ by $gRg'$ whenever $g^{-1}g'\in N$. The equivalence classes are simply the cosets $gN$. If $N=ker(f)$, then $f(g)=f(g')\iff gN=g'N$, so you have a continuous isomorphism $\tilde f:G/N\to H$ (Note that it is a homomorphism by the first isomophism theorem of groups). As $f$ is open, $\tilde f$ is a homeomorphism.