Image of Group Homomorphism is Hausdorff Implies Kernel is Closed

Theorem

Let $G$ and $H$ be topological groups.

Let $f: G \to H$ be a morphism.

Let its image $\Img f$ be Hausdorff.

Then its kernel $\map \ker f$ is closed in $G$.

Proof

By Image of Group Homomorphism is Subgroup, $\Img f$ is a group.

Let $e$ be the identity of $H$.

By Topological Group is Hausdorff iff Identity is Closed, $\set e$ is closed in $\Img f$.

Because $f$ is continuous, $\map \ker f = \map {f^{-1} } e$ is closed in $G$.

$\blacksquare$