If $G$ is profinite, and $A$ is discrete, $f: G \to A$ is continuous $\implies$ $f$ factors through a normal open subgroup

Question

Let $G$ be a profinite group; that is compact, and totally disconnected.

Take $A$ a discrete space, and a continuous map $f: G \to A$.

$\exists N$ open and normal in $G$ and a continuous map $g: G/N \to A$ with $g \circ \pi = f$ where $\pi$ is the projection

What I understand from the proof in my book so far:

Since $f$ is continuous, the image of $f$ is compact, and hence finite, since $A$ is discrete.

We can write Im$f = \{x_1,..x_n\}$, and set $X_i = f^{-1}\{x_i\}$.

Then each $X_i$ is clopen, and so it is a finite union of cosets of the form $Vg$.

The book then says to find a normal open subgroup $N$ that is contained in each such $V$, and so each $X_i$ is a union of cosets of $N$.

Can you explain this last part?

Answer 1

Let $V$ be an open neighbourhood of the identity. Then it contains an open subgroup $H$. As $G$ is compact, $H$ has finite index. Therefore it has finitely many conjugates. The intersection $K$ of the conjugates of $H$ is normal and open, so also has finite index. Therefore $V\supseteq K$ where $K$ is an open normal subgroup.

If you have finitely many $V$s, then intersect their corresponding $K$s.