Is it possible to show that $H^1(M_g)=\mathbb{R}^{2g}$ using that $H^1(X)\cong \hom(\pi_1(X),\mathbb{R})$?

Question

I want to show that if $M_g$ is a compact Riemann surface of genus $g$, then $$\hom_{\textrm{Grp}}(\pi_1(M_g),\mathbb{R})\cong \mathbb{R}^{2g}.$$ I already know that $$\pi_1(M_g)=\langle a_1,b_1,\dotsc,a_g,b_g\:|\:[a_1,b_1]\cdots[a_g,b_g]=e\rangle.$$

In the end this is just a group theory result. However I can't seem to prove it. It may help that this isomorphism is even a vector space isomorphism.

Answer 1

As $\Bbb R$ is an Abelian group, each map from $\pi_1(M)$ to $\Bbb R$ factors through the map $\pi_1(M)\to\pi_1(M)^{\text{ab}}$, the Abelianisation of $\pi_1(M)$. This is the Abelian group with the "same" presentation as $\pi_1(M)$. This means that $\pi_1(M)^{\text{ab}}$ is the free Abelian group on $2g$ generators, so isomorphic to $\Bbb Z^{2g}$. Then $$H^1(M,\Bbb R)\cong\text{Hom}(\Bbb Z^{2g},\Bbb R)\cong\Bbb R^{2g}.$$