Let $L_0$ denote the trivial complex line bundle over $S^2$ and let $L_n$ denote the complex line bundle over $S^2$ with $c_1[L_n]=n \in \Bbb Z$. Let $\Bbb H_n$ denote the $n$-th Hirzebruch surface, which is the projectivization $\Bbb P(L_n\oplus L_0)$. Thus $\Bbb H_n$ is a $S^2$-bundle over $S^2$. Let $s,f\in H_2(\Bbb H_n;\Bbb Z)$ denote the homology class of the zero section and a fiber, respectively. I want to show that $H_2(\Bbb H_n;\Bbb Z)$ is freely generated by $s$ and $f$ (without using algebraic geometry if possible).
What I've tried is as follows: Since $\Bbb H_n$ is simply-connected, $H_1(\Bbb H_n)=0$ and hence $H_3(\Bbb H_n)=0$. Decomposing the base $S^2$ to upper and lower hemispheres, we can decompose $\Bbb H_n$ into union of two copies of $S^2\times D^2$ with intersection $S^2\times S^1$. Using Mayer-Vietoris, we get $0\to H_2(S^2\times S^1)=\Bbb Z \to H_2(S^2\times D^2)\oplus H_2(S^2\times D^2)=\Bbb Z^2 \to H_2(\Bbb H_n)\to H_1(S^2\times S^1)=\Bbb Z\to 0$. Thus $\text{rank} H_2(\Bbb H_n)=2$. But what I can't show are:
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How do we know that $H_2(\Bbb H_n)$ is torsionfree?
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How do we know that $s$ and $f$ generate $H_2(\Bbb H_n)$?
Best Answer
There is a vast generalization of this due to the Leray-Hirsch theorem. Namely, if $P$ is the projectivization of a rank $n$ bundle $E \rightarrow X$ then the cohomology of $P$ as a module over $H^* (X)$ (action given by the pullback) is $H^*(X) \otimes H^*(\mathbb{C}P^n)$.
Here is a refrence written by Stephan Stolz, see pages 53-55.