Suppose $\varphi : G\to H$ is a surjective homomorphism of topological groups, then $K=ker\varphi$ is a closed normal subgroup of $G$. If $G$ is compact, then $\varphi' : G/K\to H$ is an open map.
It is straightforward to me that $K$ is a normal subgroup, could some tell me why $K$ is closed and why if $G$ is compact then $\varphi'$ is open?
Thanks!
Since $\ker\varphi=\varphi^{-1}\bigl(\{e_H\}\bigr)$, since $\varphi$ is continuous and since $\{e_H\}$ is a closed set, $\ker\varphi$ is a closed set.
If $G$ is compact, then $\varphi'$ is a continuous bijection with compact domain. Therefore, it is a homeomorphism (it maps closed sets into closed sets); in particular, it is an open map.