Hi,All:
I am seeing a result in which the following sequence, in the context of the genus-g surface Sg, is described as being exact:


1-->Tg-->$M^{(2)}g$-->$Sp^{(2)}(2g,\mathbb Z)$-->1


Where :
 i)Tg is the Torelli group ( subgroup of Mg--mapping-class group on Sg) which induces the
 identity map in homology $H\_1(Sg,\mathbb Z)$;


ii) $M^{(2)}g$ is the subgroup of Mg that induces the identity in $H\_1(Sg,\mathbb Z\_2)$


iii)$Sp^{(2)}(2g,\mathbb Z)$ is the subgroup of $Sp(2g,\mathbb Z)$ that acts trivially on


$H\_1(Sg,\mathbb Z\_2)$, i.e., if f(c)=c', then c~c' (homologous). Equivalently, 


it is also the kernel of the mod2-reduction map (map that takes a $\mathbb Z$-chain into


a $\mathbb Z\_2$-chain coefficient-by-coefficient.


$Sp(2g,\mathbb Z)$ is the subgroup of Aut$H\_1(Sg,\mathbb Z)$ that preserves the intersection


form in $H\_1(Sg,\mathbb Z)$
----------------------------


And the only non-trivial map $\Phi: M^{(2)}g$-->$Sp^{(2)}(2g,\mathbb Z)$ is the induced


map on $H\_1(Sg,\mathbb Z)$ (other maps are inclusions/projections) by $M^{(2)}g$.


Now, I get the first part: by the naturality of the mod2-reduction from $H\_1(Sg,\mathbb Z)$


to $H\_1(Sg,\mathbb Z\_2)$ , maps that induce the identity in $H\_1(Sg,\mathbb Z)$ also induce


the identity in $H\_1(Sg,\mathbb Z\_2)$, so that Tg is contained in $M^{(2)}g$, but I cannot


see why the kernel of the induced map is precisely Tg.


Any Ideas?



Sg: How to Show this Sequence is Exact?
Unless I am very confused (which is possible), $M\_g/M^{(2)}\_g$ is isomorphic to $Sp(2g, \mathbb{Z}\_2),$ pretty much by definition, whereupon your statement follows from the nine-lemma (where $M\_g$ is in the central position, and the right column has $1\rightarrow Sp^{(2)}(\mathbb{Z}) \rightarrow Sp(\mathbb{Z}) \rightarrow Sp(\mathbb{Z}\_2) \rightarrow 1.$


