Independence of Etale Fundamental Group on Base Point

Question

I'm reading Milne's notes on Etale Cohomology, where on page 27, there Remark 3.3 says that

(a) If $\overline{\overline{x}}$ is a second geometric point of $X$, then there is an isomorphism, $\pi_1(X,\overline{x})\to \pi_1(X,\overline{\overline{x}})$, well-defined up to conjugation.

While this result is not unexpected, I am not quite sure if it is immediate. I'm trying to show that the two versions of $F$ we get by considering these two different points are isomorphic, but I'm unable to do so. Moreover, what does conjugation really mean in this context? To conclude my questions are:

1. Is this remark immediate? If so, what is it that I'm missing?
2. What does conjugation mean in this context?

I'm not sure if this question has already been asked on the site previously, and I apologise if it is indeed so.

Thanks.

Answer 1

In some sense this is guaranteed by the fact that we are requiring the etale fundamental group to be profinite. To make this rigorous it's useful to use the language of Galois categories. I highly suggest [Cad] to learn about Galois categories and the relationship to etale fundamental groups.

So then, let $\mathcal{C}$ be a Galois category and let

$$F_i\colon\mathcal{C}\to\mathbf{FinSet},\qquad i=1,2$$

be two fiber functors. We define the set paths from $F_1$ to $F_2$, denote $\pi_1(\mathcal{C};F_1,F_2)$ to be the set equivalences $F_1\to F_2$.

Theorem (Grothendieck, see [Cad, Theorem 2.8 (2)): The set of paths $\pi_1(\mathcal{C};F_1,F_2)$ is always non-empty.

In other words every 'Galois category is path connected'.

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As is well-known if $X$ is a connected scheme the category $\mathbf{FEt}_X$ of finite etale covers is Galois (e.g. see [Cad, Theorem 5.10]). Moreover, for any geometric point $\overline{x}$ of $X$ the functor

$$F_{\overline{x}}\colon \mathbf{FEt}_X\to \mathbf{FinSet},\qquad (Y\to X)\mapsto |Y\times_X \overline{x}|$$

is a fiber functor. In particular, for any two geometric points $\overline{x}$ and $\overline{y}$ we can define the set of etale paths from $\overline{x}$ to $\overline{y}$ as:

$$\pi_1^\mathrm{et}(X;\overline{x},\overline{y}):=\pi_1(\mathbf{FEt}_X;F_{\overline{x}},F_{\overline{y}}).$$

We can then interpret Grothendieck's theorem as saying that 'any connected scheme is path connected (for finite etale covers)'.

But, while Grothendieck's theorem is pure abstraction (reducing essentially to the 'combinatorics' of the category of finite $G$-sets of a profinite group $G$) it would be nice to explain how one can, in practice, produce etale paths from $\overline{x}$ to $\overline{y}$. This is particularly true as more involved notions of fundamental groups/Galois categories don't enjoy the luxury of Grothendieck's theorem -- they a pirori can have non-isomorphic fiber functors and you might have to do serious work to prove the existence of 'paths' (e.g. look at [ALY], [BS], and [dJ]).

The idea for constructing etale paths is is simple: "try to find a 'simply connected' space containing $\overline{x}$ and $\overline{y}$ in its image".

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As a warm up, let's assume that $X$ is a sufficiently nice connected topological space so that the category $\mathbf{Cov}_X$ is 'like a Galois category' (more precisely a tame infinite Galois category -- see [BS]). If we have have a point $x$ of $X$ we can form a fiber functor

$$F^X_x\colon \mathbf{Cov}_X\to \mathbf{Set},\qquad (Y\to X)\mapsto |Y\times_X x|$$

If we have another point $y$ of $X$ then we should have an isomorphism $F_x\to F_y$, we should have a path from $x$ to $y$.

To see how the above idea of 'capturing $x$ and $y$ in the image of a simply connected space' helps, let us assume that $X'$ is a simply connected (again let me forego the necessary topological adjectives) and $f\colon X'\to X$ is a map such that $f(x')=x$ and $f(y')=y$. I then leave it for you to check the existence of natural isomorphisms

$$F^X_x\cong F^{X'}_{x'}\circ f^\ast,\qquad F^X_y\cong F^{X'}_{y'}\circ f^\ast$$

where $$f^\ast\colon \mathbf{Cov}_{X}\to \mathbf{Cov}_{X'},\qquad (Y\to X)\mapsto (Y\times_X X'\to X').$$

I assure you that this is easy and doesn't use the fact that $X'$ is simply connected. In fact, it shows you an important stronger claim: that to find an isomorphism $F^X_x\cong F^X_y$ it suffices to find $x$ and $y$ as the images of points $x'$ and $y'$ under map $X'\to X$ such that $F^{X'}_{x'}\cong F^{Y'}_{y'}$.

In any case, we are reduced to showing that $F^{X'}_{x'}\cong F^{X'}_{y'}$, but if we are assuming that $X'$ is simply connected, this is easy! Indeed, we know that as $X'$ is simply connected one has a (non-canonical) isomorphism over $X'$ between $Y'$ and a disjoint union of copies of $X'$. In particular, for any point $z'$ of $X'$ the map $F_{z'}^{X'}(Y') \to Y'$ induces a bijection on connected components. In particular, we get a natural diagram $$F^{X'}_{x'}(Y')\xrightarrow{\approx}\pi_0(Y')\xleftarrow{\approx}F^{X'}_{y'}(Y')$$

which produces a natural bijection $F_{x'}^{X'}(Y')\to F_{y'}^{X'}(Y')$ which is functorial in $Y'$.

Of course, the relationship with topological paths is clear: if $p\colon [0,1]\to X$ is a topological path then, as $[0,1]$ is somply connected, we obtain from the above procedure a path $F^X_x\to F^X_y$ where $x=p(0)$ and $y=p(1)$. So, you really can think of these isomorphisms as being a notion of 'generalized path'.

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So, returning to the scheme world, we see that to find etale paths $\overline{x}\to \overline{y}$ in our connected scheme $X$ it suffices to find a 'simply connected object' mapping to $X$ and containing these geometric points in their image. I leave it to you to check (it's useful to think about the stronger claim above!) that

1. by connecting any two points with a sequence of affine opens,
2. then passing to irreducible components,
3. then passing to the reduced locus (use Tag 04DZ),
4. then showing that if $\overline{x}$ and $\overline{y}$ have the same underlying point of $\mathrm{Spec}(R)$, call it $x$, then $F_{\overline{x}}\cong F_{\overline{y}}$ (hint: use our stronger claim idea above by finding an algebraically closed field containing both $k(\overline{x})$ and $k(\overline{y})$ over $k(x)$ and using the fact that algebraically closed fields are simply connected),

that it suffices to deal with the case of $\mathrm{Spec}(R)$ where $R$ is an integral domain and we may freely change the geometric point $\overline{x}$ (resp. $\overline{y}$) to any other geometric point with the same underlying point $x$ (resp. $y$).

In this case it's easy to produce such a 'simply connected space'. Indeed, let $\overline{R}$ be the integral closure of $R$ in an algebraic closure $\overline{K}$ of its field of fractions. Let $\tilde x$ and $\tilde y$ be lifts of the points $x$ and $y$ underlying $\overline{x}$ and $\overline{y}$ to $\mathrm{Spec}(\overline{R})$. The natural map

$$\tilde{x}\colon \mathrm{Spec}(k(\tilde x))\to \mathrm{Spec}(R),\qquad (\text{resp. } \tilde y\colon \mathrm{Spec}(k(\tilde y))\to \mathrm{Spec}(R))$$

is a geometric point (why are these residue fields algebraically closed?) anchored at $x$ (resp. $y$), and so we may assume they were the geometric points we started with. We are then done by noting that $\mathrm{Spec}(\overline{R})$ is simply connected which follows from Tag 0BQM.

References:

[ALY] Achinger, P., Lara, M. and Youcis, A., 2021. Geometric arcs and fundamental groups of rigid spaces. arXiv preprint arXiv:2105.05184.

[BS]Bhatt, B. and Scholze, P., 2015. THE PRO-ÉTALE TOPOLOGY FOR SCHEMES. Astérisque, 369, pp.99-201.

[Cad] Cadoret, A., 2013. Galois categories. In Arithmetic and geometry around Galois theory (pp. 171-246). Birkhäuser, Basel.

[dJ] De Jong, A.J., 1995. Étale fundamental groups of non-Archimedean analytic spaces. Compositio mathematica, 97(1-2), pp.89-118.