Let $X$ be a path-connected, locally path-connected and semilocally simply connected space and $x\in X$ an arbitrary base point. Let $p: Y\rightarrow X$ be a connected, finite cover and let $\hat{\pi}_1 (X,x)$ be the profinite completion of the fundamental group of $X$ based at $x$.

The fundamental group $\pi_1 (X,x)$ acts on the fibre $p^{-1}(x)$ via the monodromy action; in Galois Groups and Fundamental Groups (the proof of Corollary 2.3.9), Szamuely writes that

the action of $\pi_1 (X,x)$ on $p^{-1}(x)$ factors through a finite quotient, so we obtain an action of $\hat{\pi}_1 (X,x)$ as well. The stabiliser of each point $y\in Y$ [or particularly $y\in p^{-1}(x)$] is a subgroup [of $\pi_1 (X,x)$] of finite index, and hence contains a normal subgroup of finite index by the lemma. Therefore the stabiliser of $y$ under the action of $\hat{\pi}_1 (X,x)$ is an open subgroup in the profinite topology (being a union of cosets of an open normal subgroup)...

I understand all of this (I think) except for the last sentence. Here's what I understand:

Write $U_y \subseteq \pi_1 (X,x)$ for the stabiliser of $y\in p^{-1}(x)$. Then there is a normal subgroup $N$ of $\pi_1 (X,x)$ contained within $U_y$ (this is a group-theoretic lemma mentioned above) and every element of $N$ fixes $Y$, so by the Orbit-Stabiliser theorem $N$ has finite index in $\pi_1 (X,x)$. Therefore there is a natural projection $\hat{\pi}_1 (X,x)\rightarrow \pi_s (x,x)/N$ and composing this with the induced action of $\pi_1 (X,x)/N$ on $p^{-1}(x)$ gives an action of $\hat{\pi}_1 (X,x)$ on $p^{-1}(x)$.

Can anyone expand the brief explanation given in the quotation above as to why the stabiliser of $y\in p^{-1} (x)$ in $\hat{\pi}_1 (X,x)$ must be open?


1 Answer 1


The stabilizer $Stab_x$ of $x$ contains a normal subgroup $N$ of finite index. The quotient morphism $p:\hat \pi_1(X,x)\rightarrow \hat\pi_1(X,x)/N$ is continuous where $\hat\pi_1(X,x)/N$ is endowed with the discrete topology. thus $p(Stab_x)$ is open and henceceforth $Stab_x=p^{-1}(p(Stab_x))$ is open.

  • 1
    $\begingroup$ Simple and neat, thank you! $\endgroup$
    – Alex Saad
    Dec 23, 2015 at 16:41

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