I have a statement in Bott & Tu's Differential form in Algebraic topology (p.36) that I can't understand.
We say previously that a generator of $H^1(S^1)$ is a bump $1$-form on $S^1$ which gives the isomorphism $H^1(S^1)\simeq\Bbb R^1$ under integration.
It says integration but I can't see the integration in the argument where the bump $1$-form on $S^1$ is a generator of $H^1(S^1)$.
If $\alpha\in\Omega^0(U\cap V)$ is a closed $0$-form which is not the image under $\delta$ of a closed form in $\Omega^0(U)\oplus\Omega^0(V)$ (in M-V sequence), then $d^*\alpha$ will represent a generator of $H^1(S^1)$. As $\alpha$ we may take the function which is $1$ on the upper piece of $U\cap V$ and $0$ on the lower piece. Now $\alpha$ is the image of $(-\rho_V\alpha,\rho_U\alpha)$ (here, $\{\rho_U,\rho_V\}$ is a smooth partition of unity subordinate to $\{U,V\}$). Since $-d(\rho_V\alpha)$ and $d\rho_U\alpha$ agree on $U\cap V$, they represent a global form on $S^1$; this form is $d^*\alpha$. It is a bump $1$-form with support in $U\cap V$.
Where does the integration be used here?
There is a map $$ H^1(S^1) \to \mathbb{R} $$ given by $\omega \mapsto \int_{S^1} \omega$. A priori this might be the zero map, but they've proven that it isn't by constructing a closed one-form with non-zero integral, and they've also shown the one-dimensionality of the left-hand vector space. So it is an isomorphism.