In [Kaniuth, Taylor, Induced representations of locally compact groups](https://www.zbmath.org/?q=an%3A1263.22005) 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](https://en.wikipedia.org/wiki/Semidirect_product), 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.



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> Why is the inverse of this map continuous?
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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)](https://sidneymorris.net/Morris141.pdf).)


(Also [asked on math.stackexchange](https://math.stackexchange.com/questions/4479990/topological-semi-direct-products) but without a conclusive answer.)



Topological semi-direct products of groups
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.


