Group theoretic lemma about the extension of homomorphisms of profinite groups I have a question about a group-theoretic lemma proven in Galois Groups and Fundamental Groups by Tamas Szamuely. Suppose we have a profinite group $\Gamma$, a closed normal subgroup $N \subset \Gamma$ and a finite group $G$. Let $n \in N, \sigma \in \Gamma, g \in G$ . We have a left action on $Hom(N,G)$ (the set of continuous homomorphisms $N \rightarrow G$) of $G$ by conjugation, i.e. $g \cdot \phi(n) = g \phi(n) g^{-1}$ and a right action of $\Gamma$ by $(\phi \cdot \sigma)(n) = \phi(\sigma n \sigma^{-1})$. This two actions commute, $g \cdot (\phi\cdot \sigma) = (g \cdot \phi) \cdot \sigma$. 
Now, in this book there's a lemma saying that if $S \subset Hom(N,G)$ is stable under the actions of $G$ and $\Gamma$ and $G$ acts freely and transitively on $S$, then every $\phi\in S$ extends to a continuous homomorphism $\widehat{\phi}:\Gamma \rightarrow G$.
The author proves this by making a homomorphism by noting that $\forall \sigma \in \Gamma$ there exists a unique $g_\sigma \in G$ such that $\phi(\sigma n \sigma^{-1}) = g_\sigma \phi(n) g_{\sigma^{-1}}$ due to the fact that $S$ is stable under $G$ and $G$ acts freely and transitively on $S$. Naturally he puts $\widehat{\phi}(\sigma) = g_\sigma$ and he proves that this is a continuous extension of $\phi$. Now, I understand that it's an extension of $\phi$, but I don't understand why it's continuous. The author claims that it's enough to prove that $\ker(\widehat{\phi})$ is open (which I understand). He then goes on saying that $\ker(\widehat{\phi})$ consists of those $\sigma \in \Gamma$ that fix $\phi$ (which I also understand), but then he says that this is open due to the continuity of $\phi$. Why is that ?
Any help would be appreciated. 
Edit: I know that if I can prove that the right action of $\Gamma$ on $Hom(N,G)$ (with the discrete topology) is continuous, the continuity of $\widehat{\phi}$ follows since $ker(\widehat{\phi}) = Stab(\phi)$ and this is open iff the action of $\Gamma$ is continuous. So my question boils down to proving that this action is continuous.
 A: You can equally argue that $\tilde{\phi}:\Gamma\rightarrow G$ is continuous once you know that its kernel is closed in $\Gamma$, since $G$ is a finite group. Indeed, given that $\ker\tilde{\phi}$ is closed, we find $\tilde{\phi}^{-1}(g)=\gamma\ker\tilde{\phi}$ where $\gamma$ is any element in $\tilde{\phi}^{-1}(g)$. As multiplication by $\gamma$ constitutes a homeomorphism of $\Gamma$, we see that $\tilde{\phi}^{-1}(g)$ is closed as well, hence the preimage of every subset of $G$ must be closed by finiteness of $G$, which implies continuity of $\tilde{\phi}$. I will therefore prove that $\ker\tilde{\phi}$ is a closed subgroup of $\Gamma$:
To begin with, the conjugation action $$\Gamma\times N\rightarrow N,~(\gamma,n)\mapsto \gamma n\gamma^{-1}$$
is clearly continuous. Postcomposing with a continuous $\phi\in\hom(N,G)$ yields a continuous map
$$\Gamma\times N\rightarrow G,~(\gamma,n)\mapsto \phi(\gamma n\gamma^{-1}).$$
We have another continuous map
$$\Gamma\times N\rightarrow G,~(\gamma,n)\mapsto \phi(n)^{-1}$$
constructed by composing the projection $\Gamma\times N\twoheadrightarrow N$ with $\phi$ and with the inversion $G\rightarrow G,~g\mapsto g^{-1}$. Both maps yield a continuous map into $G\times G$ and therefore, by postcomposing with the multiplication map, we finally arrive at a continuous map
$$\Gamma\times N\rightarrow G,~(\gamma,n)\mapsto \phi(\gamma n \gamma^{-1})\phi(n)^{-1}.$$
Now let $A\subset \Gamma\times N$ be the preimage of $1\in G$ by the above map. As $G$ is discrete, $A$ will be both open and closed. As $\Gamma$ is profinite, points are closed, hence the sets $\Gamma\times \{n\}$ are closed for all $n\in N$. As a consequence, for any fixed $n\in N$, the set $\Gamma_n\subset \Gamma$ that is defined via
$$\Gamma_n\times\{n\}=A\cap (\Gamma\times \{n\})$$
must be closed since the projection $\Gamma\times N\twoheadrightarrow \Gamma$ is a closed map. The stabilizer of $\phi$ in $\Gamma$ can now easily be seen to equal $\bigcap_{n\in N}\Gamma_n$ and is therefore closed as well. As this is precisely the kernel of $\tilde{\phi}$, we are done.
