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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
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3 Answers 3

up vote 4 down vote accepted

(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 $\theta$.

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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.

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Can you indicate where I'm assuming that $C$ and $C\times_{X,f}X$ are isomorphic over $X$? –  oxeimon Jul 20 at 9:20
    
I've edited my original post to be a little more specific. –  oxeimon Jul 20 at 15:52
    
It follows from the assumption of the pullback diagram with the identity that $C$ and $C\times_{X,F}X$ are isomorphic because they both satisfy the universal property. The "right" pullback is the one where the map is $p^\ast f:C\to C$ - this is typically a non-trivial automorphism and this allows to see how the automorphism $f$ acts on the étale fundamental group. –  Matthias Wendt Jul 20 at 16:21

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
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Let me offer you one piece of advice, based on much experience: never ever say that two functors are "identically the same". You will fall into traps that way by not recognizing the implicit role of specific isomorphisms (or morphisms) and compatibilities thereof. It is why you got yourself completely twisted up in this case. So no, I disagree that the functors $F_x$ and $F_x \circ f^{\ast}$ are "actually identically the same": rather, there is a certain quite obvious isomorphism between them. It is exactly by paying attention to such trivial things that I unravelled your confusion. –  user52824 Jul 22 at 5:45
    
Thank you very much. I'll definitely keep that in mind. –  oxeimon Jul 22 at 8:11

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