Open morphism maps identity component onto identity component

Question

If $f:G\rightarrow H$ is open morphism of topological groups then $f(G_0)=H_0$.

So far I get that since $f$ is continuous that means $f(G_0)$ is connected and since $f$ is homomorphism that means $f(1_G)=1_H$ so we have that $f(G_0)\subset H_0$ since $H_0$ is largest connected set that contains $1_H$. Now since for every point $x$ of $G$ and every neighbourhood $U_x$ there is connected neighbourhood of it, what mean that $G_0$ is open and so is $H_0$ since open neighbourhoud $U_x$ is mapped into nonempty open set in $H_0$.

Is this correct so far and can you help me finish the proof? How to show that this map is surjective on $H_0$?

I was also thinking about bringing qutient maps into play but I wanna try to do it this way first.

Answer 1

I do not understand why the image of $G_0$ is connected since $f$ is assumed to be open and not continuous.

Since $G_0$ is open, $f(G_0)$ is open and is a neighborhood $U$ of $1_G$, by using the fact that $U_0=U\cap H_0$ is open, we deduce that $U_0$ generated $H_0$, (see the reference) and the image of $G_0$ contained $H_0$ since $f$ is a homomorphism.

https://planetmath.org/connectedtopologicalgroupisgeneratedbyanyneighborhoodofidentity