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I'm trying to understand the proof of Proposition 5.5.6 in Szamuely's Galois Groups and Fundamental Groups. The proposition relates the kernel of the map $\phi_*: \pi_1 (S', \bar{s}')\to \pi_1(S,\bar{s})$ (induced on étale fundamental groups of connected schemes by a pointed morphism $\phi: (S',\bar{s}')\to (S,\bar{s})$, where $\bar{s}, \bar{s}'$ are geometric points) to coverings of the two schemes. More specifically, it states:

Let $U'\subseteq \pi_1 (S', \bar{s}')$ be an open subgroup, and let $X'\to S'$ be the cover corresponding to the coset space $U'\backslash \pi_1 (S', \bar{s}')$. The subgroup $U'$ contains $\ker(\phi_*)$ if and only if there exists a finite étale cover $X\to S$ and an $S'$-morphism $X_i \to X'$, where $X_i$ is a connected component of $X\times_S S'$.

I am confused about two parts in the proof, essentially coming down to the same problem:

Firstly, for the "if" direction of the proof. We pick an $X\to S$ as in the statement, which corresponds to the coset space of an open subgroup $U$ of $\pi_1 (S,\bar{s})$. Szamuely then writes "by choosing an appropriate geometric base point we may identify the component $X_i \subseteq X\times_S S'$ with the coset space $U''\backslash \pi_1 (S', \bar{s}')$ for some open subgroup $U''\subseteq \pi_1 (S', \bar{s}')$. Note that $U''$ must contain $\ker(\phi_*)$ by construction."

I don't understand why $U''$ must contain $\ker(\phi_*)$. Could somebody extend on this for me?

Secondly, in the "only if" part of the proof, we use a group-theoretic lemma to obtain an open subgroup $V\subseteq \pi_1 (S,\bar{s})$ whose intersection with $\text{im}(\phi_*)$ is $\phi_* (U')$. This $V$ gives a finite étale cover $X = V\backslash \pi_1 (S,\bar{s}) \to S$, and we can pick a connected component $X_i \subseteq X\times_S S'$ to get another open subgroup $U''\subseteq \pi_1 (S', \bar{s}')$ for which $X_i$ is the coset space $X_i = U''\backslash \pi_1 (S', \bar{s}')$. Szamuely then claims that both groups $U'$ and $U''$ contain $\ker(\phi_*)$ (here $U'$ contained the kernel by assumption).

But once again, I don't understand why $U''$ must contain $\ker(\phi_*)$. What is the reason for this? I think this is something I need to see once or twice before I get the hang of using it. Many thanks!

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The answer follows from the explanations given in the paragraphs preceding 5.5.4 and 5.5.5.

When you identify $X$ with a quotient $\pi_1(S, \bar{s})/U$, you choose a point $\bar{x}$ in the geometric fiber $F_{\bar{s}}$ over $\bar{s}$, and take $U$ to be its stabilizer. The pullback cover corresponds to considering $F_{\bar{s}}$ as a $\pi_1(S', \bar{s}')$-set via$$\phi_*: \pi_1(S', \bar{s}') \to \pi_1(S, \bar{s}).$$ The connected component $X_i$ will be the cover corresponding to the orbit of $\bar{x}$ in $F_{\bar{s}}$ as a $\pi_1(S', \bar{s}')$-set. As above, this orbit can be identified with $\pi_1(S', \bar{s}')/U''$ where $U''$ is the stabilizer of $\bar{x}$ under the action of $\pi_1(S', \bar{s}')$. This action factors through the map$$\phi_*: \pi_1(S', \bar{s}') \to \pi_1(S, \bar{s}),$$so if an element is mapped to $1$ here, it will certainly fix $\bar{x}$, and hence will be in $U''$.

The other argument is similar.

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    $\begingroup$ Thank you very much for a well-written and clear answer! I'd forgotten about the identification of a subgroup $U$ as the stabiliser of a point in the geometric fiber. $\endgroup$ – Alex Saad Aug 13 '16 at 14:29

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