About naturality of Godbillon-vey class This is a problem from Lawrence Conlon's differential manifolds a first course. I do not know how to prove in the following problem
If $f: N \rightarrow M$ is transverse to $\mathcal{F}$, prove that
$$
f^*(\operatorname{gv}(\mathcal{F}))=\operatorname{gv}\left(f^{-1}(\mathcal{F})\right) .
$$
This is called the naturality of the Godbillon-Vey class.
How do you proof the naturality of the Godbillon-Vey class?
Why does it need the assumption of transversality of $f$??
 A: Naturality is easy once you showed two things.

*

*The Godbillon-Vey class is well-defined.

*The foliation $f^{-1}(\mathcal{F})$ is well defiend.

For $\mathcal{F}$ a transversely orientable foliation of codimension $q$ with normal bundle $Q$ and tangent distribution $E$. The Godbillon-Vey class construction $\operatorname{gv}(\mathcal{F}) \in H^{2 q+1}(M)$ involves arbitrary choices.
Let $\omega \in \Gamma\left(\Lambda^q(Q)\right)$ be nowhere zero. Let $\eta \in A^1(M)$ be such that $$d \omega=\eta \wedge \omega$$. It can be shown that $\eta \wedge(d \eta)^q$ is closed, thus representing a cohomology class denoted $\operatorname{gv}(\mathcal{F})$. We can choose a different non vanishing sections in $\omega^\prime \Gamma\left(\Lambda^q(Q)\right)$ and $\eta^\prime \in A^1(M)$ such that $d\eta^\prime =\eta^\prime \wedge \omega^\prime$ but one can show that
$$[\eta \wedge(d \eta)^q]=[\eta^\prime \wedge \omega^\prime].$$
This shows that $\operatorname{gv}(\mathcal{F})$ is well defined regardless of arbitrary choices.
If $f$ is transverse then there is a canonically defined foliation $f^{-1}(\mathcal{F})$ of $N$ of codimension $q$ such that $f$ carries each leaf of $f^{-1}(\mathcal{F})$ into a leaf of $\mathcal{F}$.
Assuming those things, we still need transversality for your question. Let $\omega$ be a nonvanishing element of $\Gamma\left(\Lambda^q(Q)\right)$ then $f^{*}(\omega)$ will be a non vanishing section of $\left(\Lambda^q(\hat{Q})\right)$, where $\hat{Q}$ is the normal bundle of the foliation $f^{-1}(\mathcal{F})$. This is because transversality is equivalent to $f_x^{*} : T_{f(x)}^{*}(M) \rightarrow T_x^*(N)$ is one to one on $Q_{f(x)}, \forall x \in N$. So  $f^*(\omega)$ will be a non-vanishing section of $ \left(\Lambda^q(\hat{Q})\right)$ as the exterior power of bundles is a natural operation.
The result you are trying to understand now follows from
$$df^*(\omega)=f^*(d\omega)=f^*(\eta \wedge \omega)=f^*(\omega)\wedge f^*(\eta)$$
