Lemma 10

Let

be an exact sequence of commutative locally compact groups. The kernel of $\widehat{f}$ is equal to the image of $\widehat{g}$.

The homomorphism $\widehat{f}\circ \widehat{g}$ is trivial by duality, hence the image of $\widehat{g}$ is contained in the kernel of $\widehat{f}$. Conversely, let $\chi$ be in the kernel of $\widehat{f}$. This means that Im($f$) $=$ Ker($g$) is contained in the kernel of $\chi$, hence that there exists a character $\eta$ of Im($g$) such that $\eta \circ g=\chi$. Since the inclusion of Im($g$) into K is strict, the dual restriction mapping of characters of K to Im($g$) is surjective (Lemma 9). There therefore exists a character $\beta$ of K such that $\eta$ is the restriction of $\beta$, and then $\chi =\beta \circ g=\widehat{g}(\beta )$. One concludes that the kernel of $\widehat{f}$ is contained in the image of $\widehat{g}$.

Let us now prove Theorem 4. By duality, it is enough to prove that the sequence $\widehat{K}\widehat{\longrightarrow}^g\widehat{G}\widehat{\longrightarrow}^f\widehat{H}$ is exact when the sequence $H\overset{f}{\longrightarrow}G\overset{g}{\longrightarrow}K$ is so. Now, by Lemmas 8 and 9, the morphisms $\widehat{f}$ and $\widehat{g}$ are strict, and by Lemma 10, the kernel of $\widehat{f}$ is equal to the image of $\widehat{g}$.