Remark 3

Let $f$ be a continuous homomorphism of a topological group G into a topological group $G'$, and let H be a normal subgroup of G. $f$ induces a homomorphism $g$ of the group $G/H$ onto the quotient group $f(G)/f(H)$. This homomorphism $g$ is continuous. Moreover, if $f$ is a strict morphism of G into $G'$, then $g$ is a strict morphism of $G/H$ onto $f(G)/f(H)$; for if U is open in $G/H$, and if $\varphi$ (resp. $\varphi'$) denotes the canonical mapping of G onto $G/H$ [resp. of $f(G)$ onto $f(G)/f(H)$], then we have $g(u) = \varphi'(f(\overline{\varphi}^{-1}(u)))$, and since $\overline{\varphi}^{-1}(u)$ is open in G, it follows that $g(u)$ is open in $f(G)/f(H)$, which proves our assertion.