I'm following Both-Tu book about differential topology and I wanted to understand the construction of the Thom Class of an (oriented) $n$-vector bundle $\xi\colon E \to B$ via forms. I already know the algebraic-topology way of construct if, I want to understand what's going on when working with forms.
The authors construct a global form $\psi$ on $E^0$ called the global angular form which is identified by the property that the restriction of the fibre defines a generator of $H^{n-1}(\Bbb R^n \setminus 0)$. Now via a bump form $d\rho=\rho'(r)dr$ one can claim that $d\rho \wedge \psi$ is a generator of $H^n_c(\Bbb R^n)$. Then the authors observe (page 71):
Once we have such angular form $\psi$, it is then easy to check that $\Psi=d(\rho\wedge \psi)$ is the Thom class
What I don't understand is that it seems to me that they are claiming that the Thom class is exact and therefore $0$ is cohomology, something clearly untrue. My algebraic topology background suggests me that such formula should therefore hold in $E^0$, where it is true that the Thom class is zero ($\Psi\in H^n(E,E^0)$), but the point of the bump form was to extend it to the entire $E$.
In fact at page 74 it's claimed: Although $\psi$ is defined only outside the zero section of $E$, the form $\Psi$ is a global form on $E$ since $d\rho \cong 0$ near the zero section.
I need some guidance in understanding the aforementioned equality, because it causing me a lot of confusion!
