First isomorphism theorem for topological groups Let $f$ a continuous homomorphism from a topological group $G$ onto a topological group $H$. We denote $K = Ker(f)$.
I already proved that $\overline{f}:G/K\to H$ defined by $\overline{f} (xK)=f(x)$ is an algebraic isomorphism and continuous.
Now, I suppose that $f$ is open.
How can I prove that $\overline{f}$ is a homeomorphism? First, I tried to prove it is open, but I failed. 
Any hint? Thanks.

By definition, $\bar{f}$ is the unique continuous map such that $\bar{f}\circ\pi=f$, where $\pi:G\rightarrow G/K$ is the canonical projection.
Let $V\subseteq G/K$ be open, i.e., $\pi^{-1}(V)\subseteq G$ is open. Then, because $\pi$ is surjective, $\pi(\pi^{-1}(V))=V$, so that
$\bar{f}(V)=\bar{f}(\pi(\pi^{-1}(V)))=f(\pi^{-1}(V))$.
So, since $\pi$ is continuous and $f$ is open, $\bar{f}(V)=f(\pi^{-1}(V))$ is open, so $\bar{f}$ is indeed an open map.
