Proposition 2

Let $G$ be a topological group and let $(H_\alpha)_{\alpha \in I}$ be a family of normal subgroups of $G$ such that $H_\alpha \supset H_\beta$ whenever $\alpha \leq \beta$, and which satisfy the following condition:

(AP) For each $\alpha \in I$, $H_\alpha$ is closed in $G$ and every neighbourhood of the identity element $e$ in $G$ contains one of the $H_\alpha$ (in other words, the filter base formed by the $H_\alpha$ converges to $e$).

Then the mapping $i : G \to \tilde{G} = \varprojlim G / H_\alpha$ is a strict morphism of $G$ onto $i(G)$; $\tilde{G}$ is Hausdorff and $i(G)$ is dense in $\tilde{G}$; and finally the kernel of $i$ is the closure of $\{e\}$ in $G$. If in addition one of the $H_\alpha$ is complete, then $i$ is surjective.

Clearly the $G_\alpha = G / H_\alpha$ are Hausdorff (§ 2, no. 5, Proposition 13), hence so is $\tilde{G}$ (since it is a subspace of $\prod_{\alpha \in I} G_\alpha$). The kernel $H$ of $i$ is the intersection of the $H_\alpha$ and is therefore a closed subgroup of $G$. Since each neighbourhood of $e$ contains some $H_\alpha$, it contains $H$ and therefore [§ 3, no. 1, formula (1)] $H$ is the closure of $\{e\}$. Let us next show that $i(G)$ is dense in $\tilde{G}$. Let $f_\alpha$ be the canonical mapping $\tilde{G} \to G_\alpha$, which is the restriction to $\tilde{G}$ of the projection $pr_\alpha$; $\varphi_\alpha = f_\alpha \circ i$ is the canonical mapping $G \to G/H_\alpha$. If $U$ is any non-empty open set of $\tilde{G}$, then there is an index $\alpha \in I$ and a non-empty open set $U_\alpha$ in $G_\alpha$ such that $\overline{f}_\alpha^{-1}(U_\alpha) \subset U$ (Chapter I, § 4, no. 4, Proposition 9); therefore

but since $\varphi_\alpha$ is surjective, $\overline{i}^{-1}(U)$ is not empty, and therefore $i(G) \cap U \neq \emptyset$.

To see that $i$ is a strict morphism of $G$ onto $i(G)$, consider a neighbourhood $V$ of $e$ in $G$. There is a neighbourhood $W$ of $e$ in $G$ such that $W^2 \subset V$, and an index $\alpha \in I$ such that $H_\alpha \subset W$; it follows that $V$ contains $WH_\alpha = \overline{\varphi}_\alpha(\varphi_\alpha(W)) = \overline{i}^{-1}(\overline{f}_\alpha^{-1}(\varphi_\alpha(W)))$. Since $\overline{f}_\alpha(\varphi_\alpha(W))$ is a neighbourhood of the identity element in $\tilde{G}$, the result follows.

Finally, suppose that there is an index $\gamma \in I$ such that $H_\gamma$ is complete. To show that $i$ is surjective it is enough to prove that every family $(T_\alpha)_{\alpha \in I} \in \tilde{G}$ has a non-empty intersection. Since $T_\gamma$ is obtained by translation from $H_\gamma$, it is a complete subspace of $G$ (with respect to both the right and the left uniformities). Moreover, since every neighbourhood $U$ of $e$ in $G$ contains one of the $H_\alpha$, the corresponding set $T_\alpha$ is $U_{d-}$ (or $U_{s-}$) small, and hence the set of $T_\alpha$ contained in $T_\gamma$ is a Cauchy filter base; therefore it converges in $T_\gamma$, and since the $T_\alpha$ are closed in $G$ (since they are translations of the $H_\alpha$), their intersection is not empty.

Q.E.D.