Functoriality of fundamental group via deck transformations - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:05:09Z http://mathoverflow.net/feeds/question/24241 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24241/functoriality-of-fundamental-group-via-deck-transformations Functoriality of fundamental group via deck transformations Makhalan Duff 2010-05-11T13:46:13Z 2010-05-11T15:19:21Z <h3>Problem</h3> <p>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.</p> <h3>Naive attempt</h3> <p>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.</p> <p>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 <em>really</em> 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 $X \times_Y \tilde Y$.</p> <p>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).</p> <p>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 $\pi_1(Y,y)$.</p> <p>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?</p> http://mathoverflow.net/questions/24241/functoriality-of-fundamental-group-via-deck-transformations/24248#24248 Answer by AndrĂ© Henriques for Functoriality of fundamental group via deck transformations AndrĂ© Henriques 2010-05-11T15:15:40Z 2010-05-11T15:15:40Z <p>Your spaces are not pointed, so the fundamental group is not a functor with values in groups. It's a functor with values in <em>groupoids</em>: 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.</p> <p>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).</p> <p>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 <em>up to unique isomorphism</em> by the property that it admits a map $\tilde X\to \tilde Y$ making the following square diagram commute:<br> $\tilde X\to \tilde Y$<br> $\downarrow\qquad\downarrow$<br> $X\to Y$. $\qquad\qquad$ We then let $\pi_1(f)$ be the functor sending $\tilde X\in\pi_1(X)$ to $\tilde Y\in\pi_1(Y)$.<br><br></p> <p>In topology, a concrete construction of $\tilde Y$ involves paths in $Y$ starting from the image of some point $p\in \tilde X$.<br> I don't know how to do this construction in your arithmetic context.</p> <p><hr> 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 <em>unique</em> isomorphism.</p> <p>&bull; A universal cover is uniquely defined up to isomorphism: "it's not really well defined".<br> &bull; The space $\tilde Y$ in the above diagram is uniquely defined up to unique isomorphism: "it's truly well defined".</p> http://mathoverflow.net/questions/24241/functoriality-of-fundamental-group-via-deck-transformations/24249#24249 Answer by Makhalan Duff for Functoriality of fundamental group via deck transformations Makhalan Duff 2010-05-11T15:19:21Z 2010-05-11T15:19:21Z <p>Alright, this answer is a rephrasing of all the comments:</p> <p>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.</p>