Functoriality of fundamental group via deck transformations Problem
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?
 A: 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:
$\tilde X\to \tilde Y$
$\downarrow\qquad\downarrow$
$X\to Y$. $\qquad\qquad$ 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".
A: 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.
