Topological semi-direct products of groups

Question

In Kaniuth, Taylor, Induced representations of locally compact groups on pages 9-10 it's claimed that if $G$ is a locally compact group with closed subgroups $N,H$, with $N$ normal in $G$, with $N\cap H=\{e\}$, and with $NH=G$, then $G$ is a topological semidirect product of $N$ and $H$.

We copy the algebraic proof, defining an action $\alpha_h(n) = hnh^{-1}$ which will be suitably continuous, allowing us to construct $N \rtimes_\alpha H$. The map $N \rtimes_\alpha H \rightarrow G; (n,h) \mapsto nh$ is an isomorphism of groups, and clearly continuous.

Why is the inverse of this map continuous?

You would need to show that given nets $(n_i)\subseteq N, (h_i)\subseteq H$ with $n_ih_i\rightarrow e$, then necessarily $n_i\rightarrow e, h_i\rightarrow e$. I don't see how to do this.

(Under some conditions, e.g. that $N \rtimes_\alpha H$ is $\sigma$-compact, there are open mapping theorems for locally compact groups available, which would show this. For example, see Corollary 1.7 in Hofmann, Morris, Open Mapping Theorem for Topological Groups (pdf).)

(Also asked on math.stackexchange but without a conclusive answer.)

Answer 1

Here is a counterexample:

let $K$ be an infinite compact group, and $K^\delta$ be $K$ with the discrete topology.

Let $G$ be $K^\delta\times K$, let $N$ be equal to $K^\delta\times\{0\}$ and let $H$ be the diagonal. Then both $N,H$ are discrete, $G=NH$, $N\cap H=\{1\}$. But the canonical continuous group isomorphism $N\rtimes H\to G$ is not a topological isomorphism, since $G$ is not discrete.