Background/Setup

For any connected scheme $S$, let $\text{FEt}_S$ denote the category of finite etale $S$-schemes. Let $f : X\rightarrow Y$ be a morphism of connected schemes, then for any finite etale cover $C\rightarrow Y$, we can pull it back to a finite etale cover of $X$, so after making a choice of pullbacks, we get an exact functor $f^* : \text{FEt}_Y\rightarrow\text{FEt}_X$.

Let $x,y$ be geometric points of $X,Y$ respectively, such that $f(x) = y$. Let $F_x : \text{FEt}_X\rightarrow\textbf{Sets}$ denote the fiber functor, which sends every finite etale cover $p : C\rightarrow X$ to the finite set $p^{-1}(x)$, and similarly with $F_y$. Since $f(x) = y$, the universal property of fiber products gives us a uniquely determined isomorphism of fiber functors $\eta : F_x\circ f^*\stackrel{\sim}{\longrightarrow} F_y$, and thus for any automorphism $\alpha\in\text{Aut}(F_x)$, via $\eta$ we obtain an automorphism of $F_y$. This defines a homomorphism $\pi_1(X,x) := \text{Aut}(F_x) \rightarrow \text{Aut}(F_y) =: \pi_1(Y,y)$.

My question:

This is all well and good, until I ask myself: if $X = Y$ and $f$ is an automorphism of $X$, when does $f$ induce the identity map on fundamental groups?

Specifically, fix a scheme $X$, a geometric point $x\in X$, and an automorphism $f\in\text{Aut}(X)$ such that $f(x) = x$. For any finite etale cover $p : C\rightarrow X$, we can pull it back via $f : X\rightarrow X$ to get $C\times_{X,f}X$. However, to actually define the pullback functor $f^*$, strictly speaking for any $p : C\rightarrow X$ in $\text{FEt}_X$ we really should give a concrete construction of the fiber product $C\times_{X,f}X$. In this situation, we can define this fiber product to simply be the same object $C$, where the projection map to the first component is the identity $C = C$, and the projection map to $X$ is the composition $C\stackrel{p}{\rightarrow} X\stackrel{f^{-1}}{\rightarrow} X$.

Now, given these choices of pullbacks, we should have a uniquely determined isomorphism of fiber functors $\eta : F_x\circ f^*\stackrel{\sim}{\longrightarrow} F_x$. But....for any $p : C\rightarrow X$ in $\text{FEt}_X$, we find that because $f(x) = x$ and our choice of pullbacks, that the fiber of $f^*C$ over $x$ (ie, the geometric points of $C$ which get mapped to $x$ via $C\stackrel{p}{\rightarrow}X\stackrel{f^{-1}}{\rightarrow}X$) are unequivocally the same (not just canonically isomorphic) as the fiber of $C$ over $x$ (ie, the geometric points of $C$ which get mapped to $x$ via $C\stackrel{p}{\rightarrow}X$), and so $F_x\circ f^*$ is actually equal to $F_x$, and the uniquely determined isomorphism $\eta$ is just the identity on $F_x$, which implies that the induced homomorphism $\text{Aut}(F_x)\rightarrow\text{Aut}(F_x)$ is also the identity.

This seems to show that the answer is "always".

But...I feel like this can't be right. For example, you could take $X$ to be an elliptic curve over $\mathbb{C}$, $x$ to be the point at infinity, and $f$ to be the automorphism $[-1]$. Then the induced automorphism of its topological fundamental group is nontrivial, given by inversion. Surely the same should be true of the etale fundamental group?

thanks,

• will

Background/Setup

For any connected scheme $S$, let $\text{FEt}_S$ denote the category of finite etale $S$-schemes. Let $f : X\rightarrow Y$ be a morphism of connected schemes, then for any finite etale cover $C\rightarrow Y$, we can pull it back to a finite etale cover of $X$, so after making a choice of pullbacks, we get an exact functor $f^* : \text{FEt}_Y\rightarrow\text{FEt}_X$.

Let $x,y$ be geometric points of $X,Y$ respectively, such that $f(x) = y$. Let $F_x : \text{FEt}_X\rightarrow\textbf{Sets}$ denote the fiber functor, which sends every finite etale cover $p : C\rightarrow X$ to the finite set $p^{-1}(x)$, and similarly with $F_y$. Since $f(x) = y$, the universal property of fiber products gives us a uniquely determined isomorphism of fiber functors $\eta : F_x\circ f^*\stackrel{\sim}{\longrightarrow} F_y$, and thus for any automorphism $\alpha\in\text{Aut}(F_x)$, via $\eta$ we obtain an automorphism of $F_y$. This defines a homomorphism $\pi_1(X,x) := \text{Aut}(F_x) \rightarrow \text{Aut}(F_y) =: \pi_1(Y,y)$.

My question:

This is all well and good, until I ask myself: if $X = Y$ and $f$ is an automorphism of $X$, when does $f$ induce the identity map on fundamental groups?

Specifically, fix a scheme $X$, a geometric point $x\in X$, and an automorphism $f\in\text{Aut}(X)$ such that $f(x) = x$. For any finite etale cover $p : C\rightarrow X$, we can pull it back via $f : X\rightarrow X$ to get $C\times_{X,f}X$. However, to actually define the pullback functor $f^*$, strictly speaking for any $p : C\rightarrow X$ in $\text{FEt}_X$ we really should give a concrete construction of the fiber product $C\times_{X,f}X$. In this situation, we can define this fiber product to simply be the same object $C$, where the projection map to the first component is the identity $C = C$, and the projection map to $X$ is the composition $C\stackrel{p}{\rightarrow} X\stackrel{f^{-1}}{\rightarrow} X$.

Now, given these choices of pullbacks, we should have a uniquely determined isomorphism of fiber functors $\eta : F_x\circ f^*\stackrel{\sim}{\longrightarrow} F_x$. But....for any $p : C\rightarrow X$ in $\text{FEt}_X$, we find that because $f(x) = x$ and our choice of pullbacks, that the fiber of $f^*C$ over $x$ (ie, the geometric points of $C$ which get mapped to $x$ via $C\stackrel{p}{\rightarrow}X\stackrel{f^{-1}}{\rightarrow}X$) are unequivocally the same (not just canonically isomorphic) as the fiber of $C$ over $x$ (ie, the geometric points of $C$ which get mapped to $x$ via $C\stackrel{p}{\rightarrow}X$), and so $F_x\circ f^*$ is actually equal to $F_x$, and the uniquely determined isomorphism $\eta$ is just the identity on $F_x$, which implies that the induced homomorphism $\text{Aut}(F_x)\rightarrow\text{Aut}(F_x)$ is also the identity.

This seems to show that the answer is "always".

But...I feel like this can't be right. For example, you could take $X$ to be an elliptic curve over $\mathbb{C}$, $x$ to be the point at infinity, and $f$ to be the automorphism $[-1]$. Then the induced automorphism of its topological fundamental group is nontrivial, given by inversion. Surely the same should be true of the etale fundamental group?

EDIT: To be more specific, the way that I see that $F_x\circ f^* = F_x$ is as follows. Let $pt$ denote $\text{Spec }k$ where $k$ is an algebraically closed field. Then the geometric point $x$ is given by a morphism $x : pt\rightarrow X$. For any cover $C\rightarrow X$, $F_x(C\rightarrow X)$ is defined to be the set of geometric points of $C$ over $x$. Ie, $$F_x(C\stackrel{p}{\rightarrow} X) = \text{Hom}_X(pt,C) = \{x'\in\text{Hom}(pt,C) : p\circ x' = x\}$$ Thus, $$F_x(f^*(C\stackrel{p}{\rightarrow} X)) = F_x(C\stackrel{p}{\rightarrow}X\stackrel{f^{-1}}{\rightarrow}X) = \{x'\in\text{Hom}(pt,C) : f^{-1}\circ p\circ x' = x\}$$ But of course requiring that $f^{-1}\circ p\circ x' = x$ is the same as requiring that $p\circ x' = f\circ x$, but by assumption $f\circ x = x$, so $F_x(f^*(C\rightarrow X)) = F_x(C\rightarrow X)$. Alternatively you can also get this from the universal property of the fiber product diagram $$\begin{array}{ccc} C & \stackrel{\text{id}}{\longrightarrow} & C \\ \downarrow & & \;\;\;\downarrow p \\ X & \stackrel{f}{\longrightarrow} & X \end{array}$$ where the vertical arrow on the left is the unique one making the diagram commute. Ie, its $f^{-1}\circ p$.

thanks,

• will