I was trying to prove that the singular cohomology $H^1(\Bbb{P}^1\setminus \{0,\infty\},\Bbb{Z})$ has a deRham representative $dz/z$. That is $\frac{1}{2\pi i}dz/z \in H^1(\Bbb{P}^1\setminus \{0,\infty\},\Bbb{Z})$ will generate this group (notice that by the MV sequence we have $H^1(\Bbb{P}^1\setminus \{0,\infty\},\Bbb{Z}) \cong \Bbb{Z}$ )
what I have done so far, first $dz/z$ is a well defined smooth form on $\Bbb{P}^1\setminus \{0,\infty\}$, and we know the generator of $H_1(\Bbb{P}^1\setminus \{0,\infty\},\Bbb{Z})$ should be a closed loop winding around the origin 1 time. and we know that $\frac{1}{2\pi i}\int_\gamma dz/z = 1$ for this generator $\gamma$, and the integral is independent of the representative element $\gamma \in H^1(\Bbb{P}^1\setminus \{0,\infty\},\Bbb{Z})$. Now I want to combine these observations together.
First represent $\gamma$ by the fundamental class $[\gamma] \in H^{2-1}_{\text{dR}}(\Bbb{P}^1\setminus \{0,\infty\},\Bbb{Z})$ such that :
$$\int_{\Bbb{P^1}\setminus \{0,\infty\}}[\gamma]\cup \alpha = \int_{\gamma}\alpha$$ for any 1-form $\alpha$
Second we take the Poincare duality of $[\gamma]$ gets unique 1-form $\beta$ such that $$\int_{\Bbb{P}^1\setminus \{0,\infty\},\Bbb{Z})} [\gamma]\cup \beta = \int_\gamma \beta = 1$$ and we see $\frac{1}{2\pi i}dz/z$ is this one. therefore $\frac{1}{2\pi i}dz/z$ will generate $H^1(\Bbb{P}^1\setminus \{0,\infty\},\Bbb{Z})$, is my understanding correct?