Lemma 2

Let $X \xrightarrow{f} Y \xrightarrow{g} Z$ be an exact sequence of strict morphisms of topological groups (Algebra, Chapter II, § 1, no. 4, Remark). Suppose that $X, Y, Z$ admit Hausdorff completion groups and that the identity elements $\mathscr{E} X, Y, Z$ admit countable fundamental systems of neighbourhoods. Then $\hat{X} \xrightarrow{\hat{f}} \hat{Y} \xrightarrow{\hat{g}} \hat{Z}$ is an exact sequence of strict morphisms.

Let $N_1, N_g$ be the respective kernels off and $g$; let us write

where $f_1$ is the canonical mapping $X \to X/N_f$, $f_2$ is an isomorphism of $X/N_f$ onto $N$, and $f_3$ is the canonical injection $N \to Y$. We already know that $f_2$ is an isomorphism of $(X/N_f)^*$ onto $\hat{N}_g$ and we have just recalled that $f_1$ is an injective strict morphism of $N$, to $\hat{Y}$; if we show that $\hat{f}_1$ is a surjective strict morphism, it will follow that $\hat{f}$ is a strict morphism (General Topology, Chapter III, § 2, no. 8, Remark 2). Let $g_1$ be the canonical mapping $Y \to Y/N_g$; if we show that $\hat{g}_1$ is a surjective strict morphism of the kernel $\hat{N}_g$, we shall see as above that $\hat{g}$ is a strict morphism and the sequence $\hat{X} \xrightarrow{\hat{f}} \hat{Y} \xrightarrow{\hat{g}} \hat{Z}$ will be exact. Thus we have reduced the problem to proving that, if $Y = X/N$ (where $N$ is a normal subgroup of $X$) and $f : X \to Y$ is the canonical mapping, $\hat{f}$ is a surjective strict morphism with kernel $\hat{N}$.

Let $f_0 : X_0 \to Y_0$ be the mapping which coincides with $f$ on $X_0$; as $j_X$ (resp. $j_Y$) is a surjective strict morphism of $X$ onto $X$, (resp. $Y$ onto $Y_0$), $f_0$ is a surjective strict morphism (General Topology, Chapter III, § 2, no. 8, Remark 3). Now $X_0$ and $Y_0$ are metrizable (General Topology, Chapter IX, § 3, no. 1, Proposition 1); then it follows from General Topology, Chapter IX, § 3, no. 1, Corollary 1 to Proposition 4 and Lemma 1, that $\hat{f}_0 = \hat{f}$ is a surjective strict morphism and has kernel the closure $\hat{N}_0'$ in $X$ of the kernel $N_0'$ of $f_0$. Then it will be sufficient for us to prove that $\hat{N}_0' = \hat{N}$. Now $N_0'$ obviously contains $N_0 = j_X(N)$; it will be sufficient to show that $N_0'$ is contained in the closure $\overline{N}_0$ of $N_0$ in $X$. Now,

is an open set in $X$ which does not meet $N$; as $f$ is a surjective strict morphism, $V = f(U)$ is an open set in $Y$ not containing the identity element $e'$ of $Y$ and hence not meeting the closure of $e'$; then $j_Y(V)$ does not contain the identity element of $Y_0$. But $j_Y(V) = f_0(X_0 - \overline{N}_0)$ and hence $N'_0 \subset \overline{N}_0$, which complete the proof of Lemma 2.