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?
