I'm learning something about Hirzebrunch surfaces from here. I'm probably missunderstanding something about homotopy groups of these complex surfaces. More precisely, in the link I provided the author claim that $\pi_k(H_k)=\pi_k(S^2)\times \pi_k(S^2)$, but we have a section $s_\infty\colon S^2=\mathbb{C}\mathbb{P}^1\longrightarrow H_n$, therefore it should be $\pi_k(H_n)=\pi_k(S^2)$ as $(s_\infty)_\ast\circ\pi_\ast=id_{\pi_k(S^2)}$, where $\pi\colon H_n\longrightarrow S^2 $ is the canonical projection. What am I doing wrong?


As you point out, the projection $p_k:H_k\rightarrow S^2$ has a right homotopy inverse, or section, $s_\infty$ satisfying $\pi_k\circ s_{\infty}\simeq id_{S^2}$ (note the typo in your question). However neither map is a homotopy equivalence, not a weak homotopy equivalence. In particular, $s_\infty$ is not a left homotopy inverse for the projection,

$$s_\infty\circ p_k\not\simeq id_{H_k}.$$

It is true that the induced map $s_{\infty\ast}:\pi_*(S^2)\rightarrow \pi_*(H_k)$ is monic, and exhibits $\pi_*(S^2)$ as a direct summand of $ \pi_*(H_k)$. However its complement $W$ is non-trivial. Indeed, the fibre of $p_k:H_k\rightarrow S^2$ is the projective space $\mathbb{C}P^1\cong S^2$, and if $j_k:\mathbb{C}P^1\hookrightarrow H_k$ is the fibre inclusion, then

$$p_{k*}\circ j_{k*}=(p_k\circ j_k)_*=0,$$

so the image $j_{k*}(\pi_*(\mathbb{C}P^1))\subseteq \pi_*(H_k)$ lies in the complement $W$ to $s_{\infty *}(\pi_*(S^2))$. However, since the projection $p_k$ is a fibration there is a long exact homotopy sequence


and as $p_k$ admits a section, the homomorphism $p_{k*}$ is epic in each degree. Hence by exactness, the connecting homomorphism $\Delta$ is trivial in each degree, so the long exact sequence breaks up into short exact sequences of abelian groups

$$0\rightarrow\pi_n\mathbb{C}P^1\xrightarrow{j_{k*}}\pi_nH_k\xrightarrow{p_{k*}}\pi_nS^2\rightarrow 0$$

since $S^2$ is simply connected. The induced homomorphisms $s_{\infty*}$ splits theses sequences in each degree, giving


in each degree $n\geq 1$. Now $\mathbb{C}P^1\cong S^2$, so you get the claimed result. The only reason I have written one copy of the $2$-sphere differently is to make clear which is the fibre and which is the base of the bundle projection.


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