Lemma 9

Let $f: H\rightarrow G$ be a morphism of locally compact topological groups. If the morphism $f$ is injective and strict, then $\widehat{f}$ is surjective and strict.

Suppose that $f$ is injective and strict. The morphism $\widehat{f}$ induces by passing to the quotient a morphism $q:\widehat{G}/$ Ker($\widehat{f}$)$\rightarrow \widehat{H}$. What has to be proved is that this is an isomorphism of topological groups; by duality, it is enough for this to prove that its dual $\widehat{q}$ is an isomorphism.

Let L denote the dual group of $\widehat{G}/$ Ker($\widehat{f}$) and $p:\widehat{G}\rightarrow \widehat{G}/$ Ker($\widehat{f}$) the canonical projection. We shall first prove that $\widehat{p}$ induces, by passing to subspaces, an isomorphism of L onto $f(H)$.

One has $q\circ p=\widehat{f}$, whence $\widehat{p}\circ \widehat{q}=f$. The image of $\widehat{p}$ therefore contains $f(H)$.

Since $f$ is strict, its image $f(H)$ is a locally compact subgroup of G, and is therefore closed (TG, III, p. 22, cor. 2). Let K = $G/f(H)$ and consider the exact sequence

associated (Example 1). By duality, the morphism $\widehat{f}\circ \widehat{g}$ is trivial and therefore the image of $\widehat{g}$ is contained in Ker($\widehat{f}$) $=$ Ker($p$). Thus $p\circ \widehat{g}$ is the trivial morphism and, again by duality, $g\circ \widehat{p}$ is also trivial. It follows that the image of $\widehat{p}$ is contained in the kernel of $g$, which is equal to $f(H)$. One concludes that the image of $\widehat{p}$ is equal to $f(H)$.

Moreover, since $p$ is a strict surjective morphism, the dual morphism $\widehat{p}$ is a strict injective morphism of L into G (Lemma 8). It follows that $\widehat{p}$ induces a topological group isomorphism of L onto $f(H)$. Since $\widehat{p}\circ \widehat{q}=f$, and $f$ induces an isomorphism of H onto its image $f(H)$, the morphism $\widehat{q}$ is an isomorphism.