I have been trying to understand the homotopy exact sequence for the étale fundamental group which says

$$ 1 \rightarrow \pi_1 (\bar{X},\bar{x_0})\rightarrow \pi_1 (X,x_0)\rightarrow Gal(k)\rightarrow 1 $$ is exact.

where $\bar{X}= X \times_k k_s $ and $X$ is a scheme of finite type over the field $k$ and $k_s$ is the separable closure of the field $k$.

I don't see what $\pi_1 (X,x_0)\rightarrow Gal(k) $ is, in general one expects $\pi_1$ to be a functor (for schemes) but I don't really see how.

Also if $Y \rightarrow X$ is a morphism of schemes and $X'\rightarrow X$ is an étale covering, is there a functor from coverings of $Y$ to the coverings of $X$ ?

  • $\begingroup$ Sorry, I had to edit: xkcd.com/859 $\endgroup$ – jmc Sep 11 '14 at 8:38
  • $\begingroup$ In the second $\pi_{1}$ in the exact sequence, you also need a geometric point. $\endgroup$ – jmc Sep 11 '14 at 8:39

Your second question has a negative answer because you've got some variances backwards. When $f:G \rightarrow G'$ is a map of groups, by composition with $f$ one gets a functor from the category of $G'$-sets to the category of $G$-sets. As Sawin notes, there is a functor from finite etale covers of $X$ to those of $Y$, namely base change, which computes the effect of composition with $\pi_1(Y,y) \rightarrow \pi_1(X,x)$ when $X$ and $Y$ are connected with respective geometric points $x$ and $y$ (over $x$).

Let's make this all totally down-to-earth by paying more attention to the role of the base points. Let $k(y)$ and $k(x)$ denote the respective fields at $y$ and $x$, so there is a given $k$-embedding of $k(x)$ into $k(y)$. Choose $(X',x') \rightarrow (X,x)$ a Galois pointed connected finite etale cover with $k(x') = k(x)$, so $Y' := Y \times_X X'$ is a torsor over $X'$ for $\Gamma = {\rm{Aut}}(X'/X)$ (since $X'$ is a $\Gamma$-torsor over $X$, due to connectedness of $X'$ as a finite etale cover of $X$) that is equipped with a canonical $k(y)$-point $y'$ over $y$ via $k(y) \otimes_{k(x)} k(x') \simeq k(y)$. Each connected component of $Y'$ is therefore a Galois connected finite etale cover of $Y$, with Galois group given by its stabilizer in $\Gamma$. There is a preferred connected component $U(Y',y')$ of $Y'$, namely the one containing $y'$, and its covering group over $Y$ has just been seen to be a subgroup of $\Gamma$ in a natural way. Thus, we have a natural map $$\pi_1(Y,y) \twoheadrightarrow {\rm{Aut}}(U(Y',y')/Y)^{\rm{opp}} \hookrightarrow \Gamma^{\rm{opp}}.$$ This target is canonically a quotient of $\pi_1(X,x)$, and these maps compute the composition of $\pi_1(Y,y) \rightarrow \pi_1(X,x)$ by passage to the inverse limit over all $(X',x')$.

For your first question, you have to be more precise about the base points in order to formulate an answer. Consider a geometric point $x$ of $X$ over $k_s/k$ (i.e., there is a specified $k$-embedding of $k_s$ into the field at the geometric point $x$). Let $\kappa$ be the field at $x$. An element of $\pi_1(X,x)$ is a compatible system $\{f_{X',x'}:X' \simeq X'\}$ of $X$-automorphisms of a cofinal system of $\kappa$-pointed Galois connected finite etale covers $(X',x')$ of $X$ (but $f_{X',x'}$ certainly need not preserve $x'$!). For any finite Galois extension $K/k$ you would like to describe where the composite map $\pi_1(X,x) \rightarrow {\rm{Gal}}(k_s/k) \twoheadrightarrow {\rm{Gal}}(K/k)$ carries $\{f_{X',x'}\}$.

Consider the finite etale $X$-scheme $X_K$, which may or may not be connected. The field $\kappa$ contains $k_s$ over $k$ in a specified way, hence contains $K$ over $k$ in a specified way. Thus, there is a canonical map $K \otimes_k \kappa \rightarrow \kappa$, which is to say a $\kappa$-point $u(x,K)$ of $X_K$ over $x_K = {\rm{Spec}}(K \otimes_k \kappa)$. This point lies in a unique connected component $U(x,K)$ of $X_K$. But $X_K \rightarrow X$ is a ${\rm{Gal}}(K/k)$-torsor, so by degree considerations we see that each connected component is Galois over $X$ with covering group given by its stabilizer in ${\rm{Gal}}(K/k)$. In particular, $(U(x,K), u(x,K))$ is a $\kappa$-pointed Galois connected finite etale cover of $X$ with Galois group naturally inside ${\rm{Gal}}(K/k)$. Hence, $f_{U(x,K),u(x,K)}$ arises from a unique $\sigma(X,K) \in {\rm{Gal}}(K/k)$. This element is the image of $\{f_{X',x'}\}$ in ${\rm{Gal}}(K/k)$.

We see in particular by degree considerations (and counting sizes of various finite groups) that $\pi_1(X,x) \rightarrow {\rm{Gal}}(k_s/k)$ is surjective if and only if $U(x,K) = X_K$ for all $K$, which is to say $X$ is geometrically connected over $k$.

In the special case that $X$ is normal with function field $F = k(X)$ and with $X$ geometrically connected over $k$, we can make this even more explicit. Necessarily $k$ is separably closed in $F$ by the geometric connectedness hypothesis, so the separable closure $k_s$ of $k$ in a chosen separable closure $F_s$ of $F$ makes $k_s \otimes_k F$ a field naturally inside $F_s$ over $k$. This is visibly a Galois extension of $F$ with Galois group ${\rm{Gal}}(k_s/k)$. Hence, there is a natural quotient map $${\rm{Gal}}(F_s/F) \twoheadrightarrow {\rm{Gal}}(k_s \otimes_k F/F) = {\rm{Gal}}(k_s/k).$$ Now use ${\rm{Spec}}(F_s) \rightarrow {\rm{Spec}}(F) \rightarrow X$ as the point $x$, so normality of $X$ implies that $\pi_1(X,x)$ is naturally a quotient of ${\rm{Gal}}(F_s/F)$. The above map of field Galois groups factors as the composition of ${\rm{Gal}}(F_s/F) \twoheadrightarrow \pi_1(X,x)$ with the canonical map $\pi_1(X,x) \rightarrow {\rm{Gal}}(k_s/k)$; in other words, for normal $X$ geometrically connected over $k$, the map is a refinement of the usual functoriality of absolutely Galois groups in the theory of fields (with ${\rm{Gal}}(k_s/k)$ playing the role of the opposite group of the connected pro-etale covering $X_{k_s} \rightarrow X$).

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  • $\begingroup$ will all respect, this is very explicit, but this is not very understandable ... $\endgroup$ – Niels Nov 25 '13 at 21:34
  • $\begingroup$ @Neils: Where is the first place that is not understandable, and is there an alternative method to make it explicit (including addressing the role of the base point) if one were to take the more concrete viewpoint of $\pi_1$ in terms of automorphisms of connected finite etale covers (as that is closer to the spirit of Galois theory, which is not generally taught in terms of automorphisms of fiber functors)? $\endgroup$ – user76758 Nov 30 '13 at 0:22
  • $\begingroup$ basically it is way too involved in my opinion. I would suggest to have a look at SGA1 which I find much more clear. $\endgroup$ – Niels Dec 1 '13 at 8:30
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    $\begingroup$ @Niels: When I read SGA1 as a student I worked out the above as a translation of those constructions in terms which can be expressed "at finite level", and I found the process of connecting up those two points of view to be very helpful both for understanding proofs there and when reading SGA4, Deligne's Weil II (with its Weil sheaves), etc. I'm sorry to hear that you think it is "too involved". $\endgroup$ – user76758 Dec 1 '13 at 9:29
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    $\begingroup$ @Niels: I assumed that anyone posting a question such as the above had already tried looking at the relevant parts of SGA1 and for whatever reason didn't find that sufficient on its own, so I decided to offer an alternative viewpoint which I found instructive to think through when learning these things. Of course, the OP might find my answer to not be helpful; maybe we'll find out some day. :) $\endgroup$ – user76758 Dec 17 '13 at 14:45

To your second question, the answer is yes. The functor takes $X'$ to $Y \times_X X'$.

This actually answers your first question. Given a map $f: Y \to X$ and points $y_0$ on $Y$, $x_0$ on $X$ such that $f(y_0)=x_0$, this functor is a functor from the category of finite etale covers of $X$ to finite etale covers of $Y$, hence from $\pi_1(X,x_0)$-sets to $\pi_1(Y,y_0)$-sets, which corresponds to a unique map $\pi_1(Y,y_0) \to \pi_1(X,x_0)$.

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