First isomorphism theorem for topological groups

Question

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.

Answer 1

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.

Answer 2

Hint: Let $\pi: G\to G/K$ denote projection. If $U\subset G/K$ is open then $\pi^{-1}(U)$ is open in $G$. But $f = \bar f\circ \pi$, so $f(\pi^{-1}(U)) = \ldots$