Lemma 8

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

Since $g$ is surjective, the morphism $\widehat{g}$ is injective (lemma 2 of II, p. 205). To prove that $\widehat{g}$ is a strict morphism, it is enough to prove that for every neighbourhood U of $e$ in $\widehat{K}$, there exists a neighbourhood V of $e$ in $\widehat{G}$ such that $\overset{-1}{\widehat{g}}(V)\subset U$ (lemma 2 of II, p. 200). Let U be such a neighbourhood of $e$ in $\widehat{K}$. By definition of the topology of $\widehat{K}$, there exist a compact subset X of K and a number $\varepsilon >0$ such that U contains the set of $\widehat{z}\in \widehat{K}$ which, for every $z\in X$, satisfy $|\langle \widehat{z}, z\rangle -1|< \varepsilon$. Since $g$ is strict and surjective, there exists, by TG, I, p. 80, prop. 10, a compact subset $X_0$ of G such that $g(X_0) = X$. Let V be the neighbourhood of $e$ in $\widehat{G}$ formed by the elements $\chi \in \widehat{G}$ such that,

for every $x\in X_0$, one has $|\langle \chi , x\rangle -1|< \varepsilon$. One then has $\overset{-1}{\widehat{g}}(V)\subset U$. This proves the assertion.