Prop 6.18,pg 64:The Thom class $\Phi$ on a rank $N$ oriented vector bundle $E$ can be uniquely characterized as the cohomology class in $H^n_{cv}(E)$ which restricts to the generator of $H^n_c(F)$ on each fiber $F$.

The proof wrote that

If $\Phi'$ is a coho. class in $H^n_{cv}(E)$ that restricts to a generator on each fiber then for $w \in H^n(M)$, $$ w \wedge \pi_* \Phi' = w. $$

I do not see how this holds from the definition of $\pi_*$


From Thom Isomorphism for an oriented bundle $E$ over $M$, we have isomorphism between de Rham coho. and compact vertical coho. $$ H^*(M) \rightarrow H^{*-n}_{cv}(E)$$ We define $\Phi$ the Thom class to be the image of $1$ in $H^0(M)$ under the above Thom isomorphism.

A form $w \in \Omega^*_{cv}(E)$ is locally of type (I) or (II), the map \begin{align*} (I) \quad & \pi^* \phi f(x, t_1, \ldots, t_n) dt_{i_1} \ldots d_{t_r} \mapsto 0 , \quad r< n \\ (II) \quad & \pi^* \phi f(x,t_1, \ldots, t_n) dt_1 \ldots dt_n \mapsto \phi \int_{\Bbb R^n} f(x,t_1, \ldots, t_n) dt_1 \ldots dt_n \end{align*} is our $\pi_* :\Omega^{*}_{cv}(E) \rightarrow \Omega^{*-n}(M)$. $E$ a vector bundle over $M$.

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    $\begingroup$ Can you include some more details in your question? I'm not sure which definition for a Thom class you wish to use (as it stands the stated theorem is a perfectly suitable definition). And what is $\Phi'$? And what is $\omega$? $\endgroup$ – Tyrone Nov 12 '18 at 10:55
  • $\begingroup$ I have added the details, but more can be found in the text. Tell me if things are still unclear. $\endgroup$ – CL. Nov 12 '18 at 22:52
  • $\begingroup$ I'm not sure I quite agree with the exact wording of the statement in the book, since you could always multiply such a $\Phi'$ by a non-zero real number to get another class which restricts to a generator in each fibre. I suggest working locally and studying the equation $\pi_*(\pi^*\omega\wedge \Phi')$ which appears in the book, using the description of the map $\pi_*$ to see the result. $\endgroup$ – Tyrone Nov 13 '18 at 14:24

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