[Math] homotopy exact sequence for the étale fundamental group

ag.algebraic-geometryetale-coversfundamental-group

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 $X$ ?

Best Answer

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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