Exotic automorphisms of the fundamental group of a curve? A while back, Jordan S. Ellenberg brought the following problem to my attention.
If $G$ is a residually finite group, let $\widehat G$ be its profinite completion.
Let $S$ be a closed surface of genus $g \geq 2$, and let $\pi$ be its topological fundamental group.
Let $\mathrm{Mod}(S)$ be the mapping class group of $S$.
There is homomorphism $\widehat{\mathrm{Mod}(S)} \to \mathrm{Out}(\widehat \pi)$.  Is this map surjective?
In other words, does the geometric fundamental group of the moduli space surject the outer automorphisms of the geometric fundamental group of the curve?
Edit:  As Jordan points out, the map is not surjective.  So, the question is: 
What's the closure of the image of $\mathrm{Mod}(S)$ in $\mathrm{Out}(\widehat \pi)$?
Or, more precisely, for Henry: 
As Jordan explains, there is a map $\mathrm{Out}(\widehat \pi) \to \mathrm{Sp}_{2g}(\widehat{\mathbb{Z}}) \to \widehat{\mathbb{Z}}^\star$
Is the closure of the image of $\mathrm{Mod}(S)$ in $\mathrm{Out}(\widehat \pi)$ the preimage of 1 and -1?
 A: First of all, this question came to me (or someone else with my initials) via Mark Kisin, so I can't claim credit (and for all I know it came to him from elsewhere.)
Second:  there's one obvious obstruction to surjectivity.  Namely, the map
$$\widehat{\mathrm{Mod}(S)} \to \mathrm{Sp}_{2g}(\widehat{\mathbb{Z}}) -det \to \widehat{\mathbb{Z}}^\star$$
has image $\mathbb{Z}^\star$, which is to say $\pm1$.  On the other hand,
$$\mathrm{Out}(\widehat \pi) \to \mathrm{Sp}_{2g}(\widehat{\mathbb{Z}}) -det \to \widehat{\mathbb{Z}}^\star$$
is surjective.  So the map you ask about is definitely not surjective.  The question is whether in some sense "this is the only way the map fails to be surjective."  Since I don't have a precise meaning in mind for the phrase in quotes, one might just say "what is the closure of the image of the mapping class group in $\mathrm{Out}(\widehat \pi)$?"
By the way, is there a topological proof that $\mathrm{Out}(\widehat \pi) \to \widehat{\mathbb{Z}}^\star$ is surjective?  The only proof I know is that if you write down an algebraic curve $X$ over $\mathbb{Q}$, the images of Frobenii in $\mathrm{Out}(\pi_1^{et}(X_\overline{\mathbb{Q}}))$ give you automorphisms of pihat with lots of different determinants.  Other than this I don't know how to construct a single element of $\mathrm{Out}(\widehat \pi)$ whose determinant is not $\pm1$!
A: Just a non-thought-out thought:
For any finite simple group $S$, consider the set of maps from $\pi$ to $S$ up to conjugacy. Now, $\mathrm{Out}(\widehat{\pi})$ acts on this set.  Now consider the composition of this action with the permutation character; you get a character $f(S)$ of $\mathrm{Out}(\widehat{\pi})$ valued in $\pm 1$. 
[If $S$ is $\mathbb{Z}/p\mathbb{Z}$,  I think but didn't check that the corresponding character is the "determinant composed with the quadratic residue symbol mod $p$," if that makes sense.]
I have no idea as to the image of the map $F = \prod_{S} f(S)$, but it seems plausible to me that it is uncountable. On the other hand, since $\mathrm{Out}(\pi)$ is finitely generated, the restriction of $F$
to it must have finite image.  (Slight clarification: restrict the product over $S$ to nonabelian finite simple groups, since the abelian ones provide no new information.)
