What is the Chern class of the Kernel of a projection map after taking a blowup? Let $S:= \mathbb{CP}^1 \times \mathbb{CP}^1 $ and $\tilde{S}$ the blowup of $S$ at one point. Let $a_1, a_2$ be generators for the cohomology $H^*(S, \mathbb{Z})$ and let $a_1, a_2$ and $E$ be the generators of the cohomology $H^*(\tilde{S}, \mathbb{Z})$, where $E$ is the exceptional divisor. Let $\pi_2: \tilde{S}\rightarrow \mathbb{CP}^1$ be the projection onto the second factor. 
$\textbf{Question:}$ What $c_1(Ker ~~d \pi_2)$? It has to be a linear combination of $a_1$, $a_2$ and $E$. 
Here $c_1$ denotes the first Chern class and ``Ker'' means Kernel.  
 A: Edit: This version fixes some sloppiness in the original answer.
One way to do this is the following. 
First note that if $p$ is a point in a smooth surface $X$, then $H^2(X-\{p\},\mathbf Z) = H^2(X, \mathbf Z)$ by a Mayer--Vietoris argument. Then by naturality of Chern classes, in order to calculate $c_1(V)$ for any vector bundle $V$ one can restrict $V$ to $X-\{p\}$ and work there.
Why would we want to jump through these hoops? The point is that for any $x \in \tilde{S}$ where the fibre of $\pi_2$ containing $x$ is smooth, the map $d \pi_2$ is a surjection. Now we can see geometrically where the singularities of fibres of $\pi_2$ are: in fact there is exactly one point $p$ on $S$ where the fibre is singular: namely, $p$ is the point of intersection of the two components of the fibre that contains $E$. 
So let's consider the open subset $\tilde{S}-p$. Restricting to this set, we get an exact sequence of vector bundles
$$0 \rightarrow K \rightarrow T \tilde{S} \rightarrow \pi_2^*T \mathbf P^1 \rightarrow 0$$
where $K$ is the kernel of $d \pi_2$. This gives $c_1(K) = c_1(T \tilde{S}) -\pi_2^* c_1 (T \mathbf P^1)$. In terms of your generators, this gives $c_1(K)=(2a_1+2a_2-E)-2a_2=2a_1-E$.
This calculation took place on $\tilde{S}-p$, but by the argument at the beginning, the same result is true on $\tilde{S}$ itself.
