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

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.

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 $$H^1(M,\Bbb R)\cong\text{Hom}(\Bbb Z^{2g},\Bbb R)\cong\Bbb R^{2g}.$$