I'm trying to understand this with a view towards the etale fundamental group where we can't talk about loops. What I'm missing is how the fundamental group functor should work on morphisms, without mentioning loops.

Naive attempt

Let's say we have a map $X \rightarrow Y$ (of topological spaces, schemes, what have you). Let's say $\tilde Y$ is $Y$'s universal cover (in the case of schemes, this only exists as a pro-object, and only in some cases, but for simplicity assume it exists) and $\tilde X$ is $X$'s fundamental group.

My first, naive, approach was the following: take $X \times_Y \tilde Y$. This is a cover of $X$ (etale is invariant to base change. Again $\tilde Y$ isn't really etale over $Y$ because it's not finite, but once we have the topological case down, ironing out the arithmetic details should be easy). So we have a map $\tilde X$ to $X \times_Y \tilde Y$.

Now, since $\tilde Y$ to $Y$ was Galois (- normal for the topologists; with group of deck transformations $\pi_1(Y, y)$) then so is $X \times_Y \tilde Y$ over $X$. With what group? It seems (and correct me if I'm wrong) that this will always be some quotient of $\pi_1(Y,y)$ (meaning that the group action of $\pi_1(Y,y)$ on $X \times_Y \tilde Y$ as a map $\pi_1(Y,y) \times X \times_Y \tilde Y \rightarrow (X \times_Y \tilde Y) \times_X (X \times_Y \tilde Y)$ is surjective but not nec. an immersion).

Since $\tilde X$ maps to $X \times_Y \tilde Y$, we get a natural map $\pi_1(X,x) \twoheadrightarrow Aut_X(X \times_Y \tilde Y)$, where, as we said, $Aut_X(X \times_Y \tilde Y)$ is a quotient of $\pi_1(Y,y)$.

This is not going to work. What is the right definition of how the fundamental group functor acts on morphisms, via a deck-transformations approach?

  • $\begingroup$ There are a couple problems with your treatment. First, you should do your operations on pointed spaces, since that is where fundamental groups live. Second, the fiber product is not necessarily connected (e.g., take X=Spec C, Y=Spec R). Your natural map is not in general a surjection. $\endgroup$
    – S. Carnahan
    May 11, 2010 at 14:07
  • $\begingroup$ Aha. Of course there are many problems with the naive approach (hence "naive"). For example, whenever it would be defined (remember that it only maps pi_1(X,x) to a quotient of pi_1(Y,y)), a problem in and of itself, it will be onto. This is clearly not true. For example look at the 2-cover of S^1 to itself. I'm looking for the right definition. $\endgroup$ May 11, 2010 at 14:13
  • $\begingroup$ Your question illustrates quite nicely one of the reasons for discouraging a deck-transformation approach to $\pi_1$, as I did in the post: mathoverflow.net/questions/546/… The information you want should all be there, but the puzzle at the end is especially relevant. $\endgroup$ May 11, 2010 at 14:28
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    $\begingroup$ I'm not sure that it is. To be rigorous for a second, let's say we're given (X,x)->(Y,y) (pt'd by geo. pts). Let F:Fet(X)->Sets be the geo. fiber functor, and G:Fet(Y)->Sets the same for Y. Then how do we get a morphism from Aut(F)->Aut(G)? This still seems to be unaddressed. $\endgroup$ May 11, 2010 at 14:38
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    $\begingroup$ Given any cover $Z$ of $Y$, pull it back to $Z_X$, a cover of $X$. Then $(Z_X)_x=Z_y$. So anything that acts on fibers over $x$ of covers of $X$ also acts on fibers over $y$ of covers of $Y$. The compatibility with morphisms is easy to check. $\endgroup$ May 11, 2010 at 14:49

2 Answers 2


Your spaces are not pointed, so the fundamental group is not a functor with values in groups. It's a functor with values in groupoids: The objects of $\pi_1(X)$ are the universal covers of $X$, and the morphisms of $\pi_1(X)$ are the isomorphisms between different universal covers.

For completeness, I recall that a universal cover of $X$ is a cover $\tilde X\to X$, such that $\tilde X$ is connected and simply connected (i.e. admits no non-trivial covering maps from other spaces).

Given a map $f:X \to Y$, and a universal cover $\tilde X\to X$, you get a universal cover $\tilde Y$ of $Y$ uniquely defined up to unique isomorphism by the property that it admits a map $\tilde X\to \tilde Y$ making the following square diagram commute:
$\tilde X\to \tilde Y$
$X\to Y$. $\qquad\qquad$ We then let $\pi_1(f)$ be the functor sending $\tilde X\in\pi_1(X)$ to $\tilde Y\in\pi_1(Y)$.

In topology, a concrete construction of $\tilde Y$ involves paths in $Y$ starting from the image of some point $p\in \tilde X$.
I don't know how to do this construction in your arithmetic context.

The tricky thing in the above argument is the difference between objects that are uniquely defined up to isomorphism, and objects that are uniquely defined up to unique isomorphism.

• A universal cover is uniquely defined up to isomorphism: "it's not really well defined".
• The space $\tilde Y$ in the above diagram is uniquely defined up to unique isomorphism: "it's truly well defined".

  • $\begingroup$ Allow me a not-entirely-useless nitpick. The thing that's unique up to unique isomorphism is $\tilde{Y}$ together with the map from $\tilde{X}$. Perhaps my English is faulty, but it's not enough to say that it admits such a map. In particular, I'm not sure what the value of your functor is, for example, if $X$ is a closed subspace of $Y$. $\endgroup$ May 11, 2010 at 15:33
  • $\begingroup$ Your pick-nit is entirely-useful. :-) The construction of $\tilde Y$ is as follows: Its points are equivalence classes of triples (point in $\tilde X$, point in $Y$, path in $Y$ from the image of the first point to the second point) modulo "homotopies", where a "homotopy" only lets the first and third pieces of data vary. $\endgroup$ May 12, 2010 at 14:29
  • $\begingroup$ Aha, so this is the general construction? Thanks very much. I'd never seen it before. $\endgroup$ May 13, 2010 at 23:37

Alright, this answer is a rephrasing of all the comments:

Personally I think the algebro-geometric language is completely superfluous here. This question is topological should have a completely topological answer. You start with a $\tilde Y$ to $Y$ (pt'd, whatever); take fiber product with $X$ and get a cover of $X$. Then $\pi_1(X,x)$ acts on the preimages of $x$ in each connected component of this fiber product, and therefore on all the preimages of $x$. Notice that over $x$ the fiber of this cover is naturally iso. to the fiber over $y$ in $\tilde Y$. Okay, good. So $\pi_1(X,x)$ acts on the fiber of $y$ in $\tilde Y$, so that gives a map to $\pi_1(Y,y)$. As Scott mentioned, the reason it's not onto is that $\tilde Y \times_Y X$ is not nec. connected.

  • $\begingroup$ $\pi_1(X,x)$ doesn't act on bare covers of $X$, just as abs. Galois group doesn't act on abstract sep'ble extensions (without embedding into chosen sep'ble closure). The univ. property of $(X',x')$ is: initial among pointed (not nec. connected) covers. Now $Y' \times_Y X$ has pt $(y',x)$, so defines $X' \rightarrow Y'_Y X$ with $x' \mapsto (y',x)$; i.e., $f':X' \rightarrow Y'$ over $f$ with $f'(x')=y'$. An $X$-aut. $g$ of $X'$ "is" $g(x') \in X'$. The $Y$-aut. $\pi_1(f)(g)$ of $Y'$ "is" $y' \mapsto f'(g(x'))$ [n.b.$(f'(g(x')),x)$ and $(y',x)$ may not be in same comp. of $Y' \times_Y X$!]. $\endgroup$
    – BCnrd
    May 11, 2010 at 15:52
  • $\begingroup$ Brian: He's not saying it acts on covers. Just on fibers over $x$, which is correct. Whew! These $\pi_1$ discussions are getting much more convoluted than I intended when writing that other ramble. $\endgroup$ May 11, 2010 at 15:58
  • $\begingroup$ Also note that the pair $(X',x')$ over $(X,x)$ is determined up to unique isomorphism, whereas $X'$ alone is not. Thus, although the property of $X'$ being connected and "simply connected" (i.e., having no nontrivial connected covering spaces) makes no reference to base point, there is no functorial structure with respect to $(X,x)$ until making a choice of point $x'$ over $x$; of course, any two choices are "created equal" in the sense that a unique $X$-aut. of $X'$ carries one to the other, but such a choice must be made to get a canonical structure. $\endgroup$
    – BCnrd
    May 11, 2010 at 15:59
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    $\begingroup$ @Minhyong: Ha-ha, I'm not trying to keep a secret identity; the "BCnrd" is a joke reference to my omission of vowels since I treat MO like Twitter by only writing in the restricted space of comment boxes. All I was trying to get at was that for $z$ in $Y'_y$, we don't need to pay attention to whether $(z',x)$ and $(\pi_1(f)(g)(z'),x)$ lie in the same connected component of $Y' \times_Y X$. Maybe OK for nice spaces, but could imagine a univ. cover exists for some spaces lacking a trivial "local theory" of paths; ignoring this connectedness issue is then more in the spirit of scheme case. $\endgroup$
    – BCnrd
    May 11, 2010 at 16:25
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    $\begingroup$ Well, anyways, Superman fools no one with those glasses. $\endgroup$ May 11, 2010 at 16:33

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