In self studying from Bott & Tu, I came across this problem which asks to find the class $u$ on $M$ such that $\Phi^2 = \Phi\wedge\pi^\ast u$, where $\Phi$ is the Thom class of an oriented rank 2 bundle, $E$, on $M$. Since for a rank 2 bundle, $\Phi\in H^2_{cv}(E)$, we must have $u\in H^2(M)$, in particular, we have that $u=0$ for $M$ a 1-manifold.
My first approach was to try and use the projection formula as follows $$\pi_\ast(\Phi^2) = \pi_\ast(\Phi)^2 = 1^2 = 1 = \pi_\ast (\Phi\wedge\pi^\ast u) = u\wedge\pi_\ast\Phi = u$$ where $\pi_\ast$ is integration along the fiber. However, this oviously makes no sense since $1$ is a $0$-form. After thinking about the defintion of $\pi_\ast$ I realized that while this gives a vector space isomorphism, it doesn't preserve the graded ring structure since, for example, for $\omega = \pi^\ast\phi f(\vec{x},t_1,t_2)dt_1, \eta = dt_2$, $\pi_\ast\omega = \pi_\ast\eta = 0$, but $\pi_\ast(\omega\wedge\eta) = \phi\int_{\mathbb{R}^2}f(\vec{x},t_1,t_2)dt_1 dt_2$ is generally nonzero. So all we really get from the projection formula is that $u = \pi_\ast(\Phi^2) $. Then from the explicit formula for $\Phi$ we can calculate $u$, which I get to be the Euler class $e(E)$. Can anyone verify that this is the case?
Something that supports this result is the fact that for the zero section $s$, $s^\ast\Phi = s^\ast\pi^\ast e = e$, however $s^\ast$ isn't necessarily injective as far as I could tell so the result couldn't be deduced from this observation.