# Sg: How to Show this Sequence is Exact?

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?

-

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 $1\rightarrow Sp^{(2)}(\mathbb{Z}) \rightarrow Sp(\mathbb{Z}) \rightarrow Sp(\mathbb{Z}_2) \rightarrow 1.$