topological isomorphism between a group and product of its subgroups

Question

I have stumbled upon the following question: Let $G$ be a $\sigma$-compact, locally compact Hausdorff group with $N$ and $H$ closed normal subgroups of $G$. Also $$N\cap H= \{e\}$$ and $$G=NH .$$ Then it is a known fact that $G$ is algebraically isomorphic to $$N\times H .$$ In order to have a topological isomorphism, the question is how to show the continuity of the isomorphism. If we show the continuity from $G$ to the product, we get automatically that the map is open by the open mapping theorem. The question is how to show this continuity. Your help is much appreciated.

Answer 1

It is easier in the other direction. The map

$$\varphi(n,h) = n\cdot h$$

is a group isomorphism [we do not yet know that it is an isomorphism of topological groups] $N\times H \to G$. It is continuous since it is the restriction of the multiplication on $G$ to the subset $N\times H \subset G\times G$.

Now, since $N$ and $H$ are both closed, they are locally compact [closed subspaces of locally compact (Hausdorff) spaces are locally compact]. And since $G$ is $\sigma$-compact, so are its closed subspaces, in particular $N$ and $H$. The product of two locally compact (Hausdorff) spaces is locally compact, and the product of two $\sigma$-compact spaces is $\sigma$-compact.

Hence $N\times H$ and $G$ are both locally compact and $\sigma$-compact (Hausdorff) groups, so by the open mapping theorem, it follows that $\varphi$ is open. But a continuous and open bijection is a homeomorphism, so $\varphi$ is indeed an isomorphism of topological groups.