question about the induced homomorphism of etale fundamental groups

Question

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

Answer 1

(This was meant to be a comment elaborating on Dan Petersen's answer, but got too long.) The error is that you cooked up an isomorphism between $F_x \circ f^{\ast}$ and $F_x$ and by means of that simply declared the two functors to be "equal" yet never tried to calculate how this identification interacts with automorphisms of the functor $F_x$. In other words, your arguments make no contact at all with the actual definition of the etale fundamental groups.

Let's consider an automorphism $\sigma$ of $F_x$, so for every $C \rightarrow X$ we get an automorphism $\sigma_C$ of the finite set $F_x(C) := C_x$ functorially in $C$ over $X$. On the other hand, for every $C$ we also have the automorphism $\sigma'_C := \sigma_{f^{\ast}(C)}$ of the finite set $F_x(f^{\ast}(C)) = C_{f(x)}$ which you're recognizing is $F_x(C)$ functorially in $C$. So there you go, $\{\sigma'_C\}$ is also an element of ${\rm{Aut}}(F_x)$ and $\{\sigma_C\}_{C} \mapsto \{\sigma'_C\}_C$ is the induced map on ${\rm{Aut}}(F_x) = \Pi_1(X,x)$ which you're wondering about. But does $\sigma'_C = \sigma_C$ for every $C$? In other words, does $\sigma_{f^{\ast}(C)}$ have the same effect as $\sigma_C$ via the natural bijection $(f^{\ast}(C))_x = C_x$ of the finite sets on which they act? If there were an $X$-isomorphism $f^{\ast}(C) \simeq C$ then since $F_x$ is functorial with respect to $X$-morphisms and $\sigma$ is an automorphism of $F_x$ as a functor then the answer would be "yes". However, there is no such $X$-isomorphism in general, as Dan Petersen has noted (and this is quite obvious if you recognize that $f^{\ast}(C)$ over $X$ is what you get from $C \rightarrow X$ by "applying $f$ to the coefficients of defining equations of $C$ over $X$" so to speak), so there is no argument and the conclusion is actually false (as you recognized yourself when $X$ is an elliptic curve, $x = 0$, and $f = -1$).

Your argument is predicated on thinking that you can really identify $f^{\ast}(C)$ with $C$ in a manner that has something to do with the functorialities which are present (that in turn require everything to be done "over $X$" in order to be relevant), but there is no such identification. So in the end the error is that the abstract scheme isomorphism $C \simeq f^{\ast}(C)$ is incompatible with the way in which $F_x$ is a functor, so that isomorphism has no connection to the automorphism group of $F_x$ as a functor.

Just to end with an example, suppose $X = \overline{X} - S$ for a connected Dedekind scheme $\overline{X}$ and non-empty finite set $S$ of distinct closed points $x_1, \dots, x_n \in \overline{X}$ where $n \ge 2$. Let $\overline{f}$ be an automorphism of $\overline{X}$ that permutes the $x_i$'s non-trivially and leaves invariant some geometric point $x$ of $X$. (Easy to make examples of this with curves over a field, or with integer rings of number fields, etc.) Let $f$ be the resulting restriction of $\overline{f}$ to an automorphism of $X$ (so $\overline{f}$ permutes the points in $S$), and let $C \rightarrow X$ be a connected finite etale cover whose branch locus $B \subset S$ is a non-empty proper subset (i.e., the normalization $\overline{C}$ of $\overline{X}$ in $C$ is non-etale over points in $B$ but etale over points in $S-B$). Then $f^{\ast}(C) \rightarrow X$ is a connected finite etale cover that is non-etale over all points of $\overline{f}^{-1}(B)$ but etale over $S - \overline{f}^{-1}(B)$. If there were an $X$-isomorphism $\theta:C \simeq f^{\ast}(C)$ then it would extend to an $\overline{X}$-isomorphism $\overline{C} \simeq \overline{f}^{\ast}(\overline{C})$ between the normalizations of $\overline{X}$ in each, so considering fiber structures over $S$ would force $\overline{f}^{-1}(B) = B$. Hence, any such $(C, f)$ for which $\overline{f}$ does not preserve $B$ (of which one can make many examples) admits no $\theta$.

Answer 2

The answer is that even though $C$ and $C \times_{X,f} X$ are equal as abstract schemes (with your definition of the fibered product), they are not equal as schemes over $X$ - their structure morphism is different.

Answer 3

This was meant to be a reply to user52824's answer, but got too long.

So, as I understand it now, I think the following is true:.

1. The functors $F_x$ and $F_x\circ f^*$ are actually identically the same.
2. However, the identity $F_x = F_x\circ f^*$ still gives a not necessarily trivial automorphism of $\pi_1(X,x) := \text{Aut}(F_x)$. In other words, given an automorphism $\sigma\in\text{Aut}(F_x)$, acting on $F_x$, the resulting automorphism of $F_x\circ f^*$ may be different from $\sigma$. Ie, the automorphism of $F_x\circ f^*$ induced from $\sigma$ acting on $F_x$ is the automorphism which on every $C\rightarrow X$ gives the automorphism $\sigma_{f^*C}$ on $(f^*C)_x$. When $C$ is not isomorphic to $f^*C$ over $X$, there is no morphism from $C$ to $f^*C$ in $\text{FEt}_X$, and so for any such automorphism $\sigma\in\text{Aut}(F_x)$, it may be the case that $\sigma_C\ne\sigma_{f^*C}$ as automorphisms of the finite sets $C_x = (f^*C)_x$. (they are not forced to be related in any way by functoriality since there is no morphism between $C$ and $f^*C$ in $\text{FEt}_X$).

Thanks for your answer. I wrote this mostly for my own benefit, but I just accepted your answer.

• will