Understanding the Whitney sum bundle - Pull backs Denote a manifold by $M$, and the tangent bundle on this by $\tau_M$, the normal bundle by $\nu_M$.
We can consider the whitney sum of these bundles denoted here: $\tau_M\oplus \nu_M$.
This should be obtained as:
$$\begin{matrix}E_1\oplus E_2&\longrightarrow&E_1\times E_2\\\downarrow&&\downarrow\\B&\stackrel{d}{\longrightarrow}&B\times B\end{matrix}$$
Such that the projection for $\tau_M\times\nu_M$ is given by $$\pi_1\times \pi_2:E_1\times E_2\to B\times B$$ gives us the induced projection from $E_1\oplus E_2$ i.e. $\pi_3:E_1\oplus E_2\to B$ where $$E_1\oplus E_2= \{(b,(e_1,e_2)):d(b)=(b,b)=(\pi_1(e_1),\pi_2(e_2))\}$$ and thus $$\pi_3:(b,(e_1,e_2))\mapsto b.$$

Does this give us that $\pi_3^{-1}(b)=\{b\}\times\pi_1^{-1}(b)\times \pi_2^{-1}(b)$?

 A: Yes, it does give you $\pi_3^{-1}(b) = \{b\} \times \pi_1^{-1}(b) \times \pi_2^{-1}(b)$. If you want you can do it by double inclusion:


*

*Suppose that $\xi = (b, e_1, e_2) \in \pi_3^{-1}(b)$, then by definition $(\pi_1(e_1), \pi_2(e_2)) = d(b) = (b,b)$, i.e. $\pi_1(e_1) = \pi_2(e_2) = b$, in other words $e_1 \in \pi_1^{-1}(b)$ and $e_2 \in \pi_2^{-1}(b)$. It follows that $\xi \in \{b\} \times \pi_1^{-1}(b) \times \pi_2^{-1}(b)$.

*Conversely suppose that $\xi = (b, e_1, e_2) \in \{b\} \times \pi_1^{-1}(b) \times \pi_2^{-1}(b)$. Then you do have $(\pi_1(e_1), \pi_2(e_2)) = (b,b) = d(b)$ and so $\xi \in E_1 \oplus E_2$, and of course $\pi_3(\xi) = b$.
In the future, you can note that it's easier to work with the following description of $E_1 \oplus E_2$:
$$E_1 \oplus E_2 = \{ (e_1, e_2) \in E_1 \times E_2 \mid \pi_1(e_1) = \pi_2(e_2) \}$$
with the projection $\pi_3$ given by $\pi_3(e_1, e_2) = \pi_1(e_1) = \pi_2(e_2)$. Roughly speaking, $E_1 \oplus E_2$ is the bundle over $B$ with fibers the products of the fibers of $E_1$ and $E_2$.
