# Why do these open subgroups of the étale fundamental group contain the kernel of an induced homomorphism?

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!

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 $U''$.
• 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. – Alex Saad Aug 13 '16 at 14:29