A differential form with integration $1$ has to be integral?

Question

Let $X$ be a compact connected Riemann surface. If $\Omega\in H^2(X,\mathbb{C})$, s.t. $\int_X\Omega=1$, then why is $\Omega$ integral?

A $d$-closed differential form $\phi$ on $X$ is said to be integral if its cohomology class in the de Rham group, $[\phi]\in H^*(X,\mathbb{C})$, is in the image of the natural mapping: $H^*(X,\mathbb{Z})\to H^*(X,\mathbb{C})$.

Answer 1

How do you give a (natural) isomorphism $H^2(X,\Bbb C)\cong\Bbb C$? You evaluate on the fundamental class $[X]\in H_2(X,\Bbb Z)$, which means—in the deRham picture—that you integrate over $X$. The integral cohomology $H^2(X,\Bbb Z)$ then corresponds to $\Bbb Z\subset\Bbb C$: A cohomology class gives you an integer for the integral if and only if it comes from an element of $H^2(X,\Bbb Z)$.