Prop 12.8 in Bott & Tu This proposition in Bott & Tu have been haunting me for a year or so since I always have to come back to this book for references. More precisely, the second equality in Proposition 12.8 in page 135 of Differential Forms in Algebraic Topology has a weird assertion.
Let $E \twoheadrightarrow X$ be a vector bundle over a manifold $X$, $S_0$ the image of the zero section, $S$ the image of a section transversal to $S_0$, $Z = S \cap S_0$, $x \in Z$ (by identifying $S_0$ and $X$) and $S_x = (N_{Z/S})_x$.
Let $\Phi$ be the Thom class of $N_{Z/X}$. The authors claim that $$\int_{S_x} \Phi = \int_{E_x} \Phi$$, because $S_x$ and $E_x$ are homotopic modulo the region in $E$ where $\Phi$ is zero.
In this context, this justification makes no sense. The unique possible theorem that they're alluding to is contained in the answer of invariance of integrals for homotopy equivalent spaces . However this still makes no sense in the equality above even by fixing a homotopy equivalence $f: S_x \rightarrow E_x$ since $f^* \Phi \neq \Phi$ might happens.
I would like a clarification of the equality mentioned above.
Thanks in advance.
 A: Let $z \in Z$. It's enough to prove the statement for a trivial bundle since the argument is done pointwise. In this case, $E = \mathbb{R}^m \times \mathbb{R}^n$, $X = \mathbb{R}^m$, $$(x,s(x)): X \rightarrow E$$, $Z = \mathbb{R}^{m - n} \subset \mathbb{R}^n$ and $$\Phi = \rho (x_1, …, x_{m + n}) dx_1 \wedge … dx_{m + n}$$, where $$\rho : E = \mathbb{R}^m \times \mathbb{R}^n  \rightarrow \mathbb{R}$$ is a bump function along the fibers of $E$, i.e., $\forall y \forall z (\rho (x, y) = \rho (x', y))$ and $\rho (x, -) : \mathbb{R}^n \rightarrow \mathbb{R}$ is a bump function.
By the differentiability of $s$, there's a $\varepsilon > 0$ such that $$C = S \cap \prod_{i = 1}^{i = m} (z_i -\varepsilon, z_i + \varepsilon) \times \prod_{i = m + 1}^{m + n} (-\varepsilon, \varepsilon)$$ is diffeomorphic $Z \times \mathbb{R}^n$ by restricting the projection $f: \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb {R}^n$ to $C$. 
Let $$supp (\rho) \subset \mathbb{R}^m \times \prod_{i = m + 1}^{m + n} (-\varepsilon, \varepsilon)$$. Then $$df_s (v_1, v_2) =(0, v_2)$$, where the tangent space at each point is being embedded in $E$ (think about a vector attached to a point in $\mathbb{R}^m \times \mathbb{R}^n$). Hence $$f^{*}\Phi = \Phi$$ holds on $\prod (z_i -\varepsilon, z_i + \varepsilon) \times \prod_{i = m + 1}^{m + n} (-\varepsilon, \varepsilon)$.
Therefore $$\int_{S_z} \Phi = \int_{S_z} f^{*}\Phi = \int_{E_z} \Phi$$.
Obs: Notice that the existence of a transversal section implies that $m \geq n$. I've kept forgetting this and it confused my intuition sometimes. This is why I'm adding this remark.
