Proposition 3

Suppose that $G$ is Hausdorff and that the $H_\alpha$ are compact and satisfy (AP).

a) *Let $L$ be a closed subgroup of $G$; then, for each $\alpha \in I$, the subgroup $L_\alpha = f_\alpha(L)$ of $G_\alpha = G/H_\alpha$ is closed, and the isomorphism $i$ of $G$ onto $\varprojlim G_\alpha$ gives by restriction an isomorphism of $L$ onto $\varprojlim L_\alpha$. If also $L$ is normal in $G$, then $L_\alpha$ is normal in $G_\alpha$ for each $\alpha \in I$, and by passing to the quotients, $i$ induces an isomorphism of $G/L$ onto $\varprojlim G_\alpha/L_\alpha$.

b) Conversely, for each $\alpha \in I$ let $L_\alpha$ be a closed subgroup of $G_\alpha$, such that $L_\alpha = f_{\alpha\beta}(L_\alpha)$ whenever $\alpha \leq \beta$. Then there is a unique closed subgroup $L$ of $G$ such that $L_\alpha = f_\alpha(L)$ for each $\alpha \in I$; and if in addition $L_\alpha$ is normal in $G_\alpha$ for each $\alpha \in I$, then $L$ is normal in $G$.

a) Since $H_\alpha$ is compact, $LH_\alpha$ is closed in $G$ ($\S$ 4, no. 1, Proposition 1, Corollary 1) and therefore $L_\alpha$ is closed in $G_\alpha$. Since $i$ identifies the topological groups $G$ and $\varprojlim G_\alpha$, and since $\varprojlim L_\alpha$ may be identified with a (topological) subgroup of $\varprojlim G_\alpha$, $i$ identifies the subgroup $\bigcap_\alpha LH_\alpha$ of $G$ with $\varprojlim L_\alpha$, and to prove the first assertion it is enough to remark that $L = \bigcap_\alpha LH_\alpha$ by Lemma 1. On the other hand, if $L$ is normal, then for each $\alpha \in I$ the mapping $f'_\alpha : G/L \to G_\alpha/L_\alpha$ induced by $f_\alpha$ is a surjective strict morphism ($\S$ 2, no. 8, Remark 3), whose kernel is the compact normal subgroup $H_\alpha L/L$ of $G/L$, the canonical image of the compact subgroup $H_\alpha$ of $G$. Since the subgroups $H_\alpha L/L$ of $G/L$ satisfy condition (AP) (§ 2, no. 8, Proposition 24), and since $G/L$ is Hausdorff, the last assertion of a) is a consequence of Proposition 2.

b) Let $f'_{\alpha \beta}$ be the restriction of $f_{\alpha \beta}$ to $L_\beta$ ($\alpha \leq \beta$). Then $(L_\alpha, f'_{\alpha \beta})$ is an inverse system of topological groups, whose inverse limit $L$ can be identified with the subgroup $G \cap \prod_\alpha L_\alpha$ of $G$. By hypothesis, $f'_{\alpha \beta}$ is surjective and its kernel is the compact subgroup $f_\beta(H_\alpha) \cap L_\beta$ of $L_\beta$; consequently (Chapter I, § 9, no. 6, Corollary 1 to Proposition 8) we have $L_\alpha = f_\alpha(L)$ for all $\alpha \in I$. If $L'$ is another closed subgroup of $G$ such that $f_\alpha(L') = L_\alpha$ for all $\alpha \in I$, then $L'H_\alpha = \overline{f'}_\alpha(L_\alpha)$, whence (Lemma 1) $L' = \bigcap_\alpha L'H_\alpha = \bigcap_\alpha \overline{f'}_\alpha(L_\alpha) = L$. Finally, the last assertion of b) follows from the formula $L = \bigcap_\alpha \overline{f'}_\alpha(L_\alpha)$, since the $\overline{f'}_\alpha(L_\alpha)$ are now normal subgroups of $G$.