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I have a problem with Proposition (1.2.1) from J. Wilson's book 'Profinite Groups'

The proposition is the following:

Let $(G, \varphi_i : G \to G_i)$ be an inverse limit of an inverse system of compact Hausdorff topological groups $(G_{i}, \varphi_{i,j} : G_j \to G_i)$ (over the directed poset $(I, \preceq)$) and let $N$ be an open normal subgroup of $G$. Then there exists $i \in I$ such that $\ker \varphi_i \leq N$. Consequently there are $N_i, H_i \leq G_i$ with $N_i \unlhd H_i$ such that $G/N \cong H_i/N_i$ as topological groups.

We already know that for such an inverse system the collection $\left\{{\varphi_j}^{-1}(U) \mid U \textrm{ open in } G_j,\, j \in I\right\}$ is a base for the topological space $G$ (which is compact and Hausdorff, too), so proving that $\ker \varphi_i \leq N$ for some $i \in I$ is easy. Then, using the third isomorphism theorem for topological groups, we can state that $G/N \cong (G/\ker \varphi_i)/(N/\ker \varphi_i)$ as topological groups, and that's ok.

Then the author states that, by the first isomorphism theorem for topological groups, we have $G/\ker \varphi_i \cong \varphi_i(G)$ as topological groups, so the result follows. Here's the first isomorphism theorem for topological groups:

Let $G, H$ be topological groups and $f : G \to H$ a continuous group homomorphism. Then $G/\ker \varphi_i \cong \varphi_i(G)$ as topological groups if and only if $f$ is open onto its image (i.e. for each open subset $U$ of $G$, the set $f(U)$ is open in $f(G)$ seen as a subspace of $H$).

So, it seems that I have to prove that $\varphi_i$ is open onto its image, before I could state that $G/\ker \varphi_i \cong \varphi_i(G)$ as topological groups. Can anyone suggest me how to do that, please? Thanks in advance...

Further details:

this proposition is stated before any result explicitly concerning pro-$\mathcal C$ (hence also profinite) groups (and it is actually used, at least in this text, to prove the first main result about them, giving some characterisation of pro-$\mathcal C$ groups), so I cannot use them.

Instead, it is possible to assume that $(G, \varphi_i)$ is the "usual" inverse limit of $(G_{i}, \varphi_{i,j})$, that is $$G = \left\{ g = (g_i) \in \prod_{i \in I} G_i \mid \forall i, j \in I \textrm{ with } i \preceq j : \varphi_{i,j}(g_j) = g_i\right\},$$ $$\varphi_i = {p_i}_{|G},$$ where the $p_i$ are the projection maps of $\prod_{i \in I} G_i$.

  • $\begingroup$ $G/\mathrm{ker} \varphi_i$ is compact in the quotient topology, $\varphi_i(G)$ is Hausdorff in the relative topology and the bijection from the quotient to the image is continuous. Hence its inverse is continuous, too. $\endgroup$ – Jochen Wengenroth May 22 '17 at 6:33
  • $\begingroup$ You're right, that's a better way to obtain what I needed... Many thanks! $\endgroup$ – FrankMiller May 22 '17 at 10:10

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