By Baer's (I believe) theorem, the mapping class group of a closed surface is actually the outer automorphism group of the fundamental group. On the other hand, the action of this group on $H_1(S, \mathbb{Z}))$ is symplectic (preserves the symplectic form, and the image under the Torelli map is $SP(2g, Sp(2g, \mathbb{Z}).$ mathbb{Z})).$On the other hand, the action of the outer automorphism group of$F_{2g}$on its abelianization is the whole$SL(2g, \mathbb{Z})$of which$SP(2g, Sp(2g, \mathbb{Z})$is a proper subgroup. So, since$Out(\pi_1(S_{g})) \neq Out(F_{2g})$the groups themselves are not isomorphic. 1 By Baer's (I believe) theorem, the mapping class group of a closed surface is actually the outer automorphism group of the fundamental group. On the other hand, the action of this group on$H_1(S, \mathbb{Z}))$is symplectic (preserves the symplectic form, and the image under the Torelli map is$SP(2g, \mathbb{Z}).$On the other hand, the action of the outer automorphism group of$F_{2g}$on its abelianization is the whole$SL(2g, \mathbb{Z})$of which$SP(2g, \mathbb{Z})$is a proper subgroup. So, since$Out(\pi_1(S_{g})) \neq Out(F_{2g})\$ the groups themselves are not isomorphic.