TITLE: Exotic automorphisms of the fundamental group of a curve?
QUESTION [10 upvotes]: 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?

REPLY [3 votes]: 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.)