# Why $H_{dR}^1(M) \simeq \mathbb R^n$ when $H_1(M,\mathbb Z)$ has $n$ generators?

Let $M$ be compact, connected and orientable manifold without boundary. Let $\sigma_1,\ldots, \sigma_n$ be generators of $H_1(M,\mathbb Z)$. Why is the map $I:H_{dR}^1(M) \to \mathbb R^n$ given by $$I([a]) = \left(\int_{\sigma_1} a , \ldots, \int_{\sigma_n} a\right)$$ an isomorphism?

I believe the proof involves Hodge theorem but I do not know where.

Observation: $H_{dR}^1(M)$ is the first De Rham cohomology group and $H_1(M,\mathbb Z)$ is the first singular homology group with coefficients on $\mathbb Z$.

• This is not true in the presence of torsion: Consider $\mathbb{RP^2}$. – Thomas Rot Feb 22 '18 at 14:55
• Ah ok, I forgot to add that $M$ is orientable. – Hugocito Feb 22 '18 at 15:09
• This is also false on $RP^3$ I think. – Thomas Rot Feb 22 '18 at 15:27