I apologize for the vague question title, but I can't think of a better subject line.

Let $f : X\rightarrow S$ be a morphism of connected schemes. Let $x,y\in X$, $s\in S$ be geometric points with $f(x) = f(y) = s$.

Let $FEt_S$ be the category of finite etale schemes over $S$, and similarly with $X$. We have fiber functors $F_s : FEt_S\rightarrow Sets$, and $F_x,F_y : FEt_X\rightarrow Sets$, and $\pi_1(X,x),\pi_1(X,y),\pi_1(S,s)$ are just the automorphism groups of the associated fiber functors.

There are canonical isomorphisms $$\beta_x : F_x\circ f^*\stackrel{\sim}{\longrightarrow}F_s\qquad\text{and}\qquad \beta_y : F_y\circ f^* \stackrel{\sim}{\longrightarrow}F_s$$

Any isomorphism $\alpha : F_x\stackrel{\sim}{\longrightarrow} F_y$ induces an automorphism $$\overline{\alpha} := \beta_y\circ\alpha\circ\beta_x^{-1}\in Aut(F_s) = \pi_1(S,s)$$

Suppose the induced map $f_* : \pi_1(X,x)\rightarrow \pi_1(S,s)$ is trivial (sends everything to the identity), and similarly $f_* : \pi_1(X,y)\rightarrow\pi_1(S,s)$, then I believe I can prove that $\overline{\alpha}$ does not depend the choice of $\alpha$. However, I'm having a lot of trouble arguing that $\overline{\alpha}$ should always be $id\in Aut(F_s)$.

Is this even true? (I mean, surely it is right. "What else could it be?") How would one prove something like this?

EDIT 1: The specific case I'm interested in is where $X\rightarrow S$ actually factors through the geometric point $s$.

EDIT 2: Okay it seems that this additional assumption makes the statement pretty easy to prove, though I would be interested in a counterexample to the original question.


1 Answer 1


I don't think that $\bar\alpha$ is always the identity. You should think of $\alpha$ as a ``path'' from $x$ to $y$, so in principle it is possible that the loop obtained as the image of $\alpha$ is not nullhomotopic, even if the image of every loop is. In fact, if the map $f$ is the universal covering space you can get every possible element in $\pi_1(S,s)$ by choosing appropriately $x$ and $y$. To support this let me try to get you an example where $\bar \alpha$ is not in the image of $f_*$ (although $f_*$ is not trivial).

Fix the ground field $\mathbb{C}$. Take $f:\mathbb{A}^1\smallsetminus \{0\}\to \mathbb{A}^1\smallsetminus \{0\}$ be the square map, and take $x=1$ and $y=-1$. Let the isomorphism $\alpha: F_x\xrightarrow{\sim} F_y$ to be obtained by "sliding" the points of the fiber along the half unit circle in the positive semiplane. Then $\bar\alpha$ should be the generator of $\pi_1(\mathbb{A}^1\smallsetminus\{0\})=\hat{\mathbb{Z}}$, while the image of $f_*$ is the index 2 subgroup.

  • $\begingroup$ So the case that I'm specifically interested in is where the map $X\rightarrow S$ factors through a geometric point, in which case topologically speaking any path in $X$ maps to the constant loop. I'll edit the question to reflect this. $\endgroup$
    – Will Chen
    Feb 28, 2016 at 4:31
  • $\begingroup$ @oxeimon Well, in that case $\bar\alpha$ is in the image of $\pi_1(\{s\},s)=*$, so there's not much to prove :). My problem with writing a counterexample is that I don't know any maps such that $f_*$ is trivial, I believe that almost any of them is a counterexample. $\endgroup$ Feb 28, 2016 at 12:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .