My questions arise on page xiv of Stix's Rational Points and Arithmetic of Fundamental Groups. Here is an excerpt:

Given a geometrically connected variety $X/k$, a fixed separable closure $\bar{k}/k$, $\bar{X} = X\times_k\bar{k}$, and $\bar{x} \in \bar{X}$ a geometric point, there is the homotopy exact sequence $$1\to\pi_1(\bar{X},\bar{x})\to\pi_1(X,\bar{x})\to \mathrm{Gal}(\bar{k}/k)\to 1$$ If $a\in X(k)$ is a rational point, there is an induced section $s_a: \mathrm{Gal}(\bar{k}/k)\to \pi_1(X,\bar{a})$, where $\bar{a}$ is a geometric point compatible with the choice of $\bar{k}/k$.

An etale path $\gamma$ from $\bar{a}$ to $\bar{x}$ on $\bar{X}$ defines an isomorphism $\gamma (-) \gamma^{-1}: \pi_1(X,\bar{a})\to\pi_1(X,\bar{x})$ compatible with the projections $\mathrm{pr}_*:\pi_1(X,\bar{x})\to \mathrm{Gal}(\bar{k}/k)$.

Then the composition $\gamma(-)\gamma^{-1}\circ s_a$ defines a section of $\mathrm{pr}_*:\pi_1(X,\bar{x})\to \mathrm{Gal}(\bar{k}/k)$, or a splitting of the above exact sequence.

Changing the etale path $\gamma$ on $\bar{X}$ varies the section over $\pi_1(\bar{X},\bar{x})$-conjugacy classes of splittings/sections. Denote the $\pi_1(\bar{X},\bar{x})$-conjugacy classes of sections of the exact sequence by $\mathscr{S}_{\pi_1(X/k)}$

Changing the base point $\bar{x}$ leads to another description of $\mathscr{S}_{\pi_1(X/k)}$ with a canonical identification between the two descriptions, which moreover satisfies the cocycle relation for composing the identifications between three choices of base points.

My questions are:

1) An etale path from $\bar{a}$ to $\bar{x}$ means an isomorphism between their fiber functors. What does it mean to be an etale path "on $\bar{X}$"?

2) Why is the isomorphism $\gamma(-)\gamma^{-1}$ compatible with the projections $\mathrm{pr}_*$?

3) Why does changing the etale path only vary the section by $\pi_1(\bar{X},\bar{x})$-conjugacy, and not $\pi_1(X,\bar{x})$-conjugacy?

4) Could someone please explain the last paragraph about the different descriptions of $\mathscr{S}_{\pi_1(X/k)}$ in a more explicit way?

Thank you.


This isn't a complete answer, but it is too long for a comment. Here are a couple of ideas relating to your first two questions that I hope you (or anyone else learning this area!) may find useful. Throughout, to better fit your notation, I will assume that $k$ is perfect.

Firstly, an etale path on $\bar{X}$ ought to mean exactly the same thing as on any scheme: an isomorphism between two fibre functors $\mathsf{FEt}_{\bar{X}}\to \mathsf{FSet}$ (denoting the categories of finite etale covers of $\bar X$ and of finite sets respectively). Quite generally, given any morphism of schemes $f: Y\to Z$, base change by $f$ gives a functor $B = B_f: \mathsf{FEt}_Z\to\mathsf{FEt}_Y$. Given two fibre functors $F_1, F_2$ on $\mathsf{FEt}_Y$ we can compose these with $B$ like this:


An isomorphism between $F_1$ and $F_2$ can be sent to an isomorphism of the composite functors. So any etale path $\gamma: F_1\to F_2$ on $Y$ maps to an etale path $f(\gamma): F_1 \circ B\to F_2\circ B$ on $Z$. Thus etale paths on $\bar{X}$ map to etale paths on $X$ under the morphism $\bar{X}\to X$.

I have also struggled recently understanding why "conjugation by an etale path $\gamma$" (on $X$, but take the image of a path on $\bar X$ if you want) should be compatible with the projection $pr_*: \pi_1 (X, \bar{x})\to \Gamma:=\text{Gal}(\bar{k}/k)$. Fortunately there seems to be a way to get around this.

Let me denote the isomorphism $\gamma (-)\gamma^{-1}:\pi_1 (x,\bar{a})\to\pi_1 (X,\bar{x})$ by $c_\gamma$. The two composite fibre functors

$$F_{\bar{a}}\circ B, F_{\bar{x}}\circ B:\mathsf{FEt}_{\text{Spec}(k)}\to\mathsf{FEt}_X\to\mathsf{FSet}$$

are both isomorphic to $F_{\bar{y}}$, where $\bar{y}:\text{Spec}(\bar{k})\to\text{Spec}(k)$ is a geometric point corresponding to the choice of fixed separable ($=$ algebraic) closure. As above, hitting these composite functors with the natural transformation $\gamma$ gives us an etale path on $\text{Spec}(k)$,

$$\alpha = pr (\gamma): F_{\bar{y}}\xrightarrow{\sim}F_{\bar{y}},$$

which has the same start and endpoints i.e. an element of its fundamental group $\Gamma$. Geometrically what was once a "path" on $X$ has mapped down to a "loop" on $\text{Spec}(k)$ under $pr$. This loop induces an isomorphism $c_\alpha: \Gamma\xrightarrow{\sim}\Gamma$ via conjugation. We then have a diagram of the form

$$\require{AMScd} \begin{CD} \pi_1 (X,\bar{a}) @>{c_\gamma}>> \pi_1 (X,\bar{x})\\ @V{pr_{*,a}}VV @VV{pr_{*,x}}V\\ \Gamma @>>{c_\alpha}> \Gamma \end{CD}.$$

(Note the vertical maps are dependent on the basepoints - the rightmost vertical map is $pr_*$ in your question.) We know $s_a$ is a section of the left vertical map. Define

$$\sigma_a = c_\gamma \circ s_a \circ c_\alpha^{-1}$$

This $\sigma_a$ is then a section of the right vertical map. Hence the rational point $a\in X(k)$ does indeed induce a section $\sigma_a : \Gamma\to\pi_1 (X,\bar{x})$ of the projection map $pr_*$ that is compatible with this difference of basepoints.


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