# Homology class of an inclusion

Let $S^2\to E \to B$ be a smooth fiber bundle with $E$ 1-connected and $B$ 3-connected. Let $i:S^2\to E$ be the inclusion of a fiber.

I have to determine the homology class $[i]\in H_2(E;\mathbb{Z})$ to which the inclusion map $i$ belongs. Note that $i$ is an embedding because each fiber is an embedded submanifold of $E$.

Since $E$ is 1-connected, the Hurewicz theorem gives an isomorphism $\pi_2(E)\cong H_2(E;\mathbb{Z})$. Also, any element $\beta\in H_2(E;\mathbb{Z})$ can be represented by an embedding $f_{\beta}:S^2\to E$.

Moreover, the LES of homotopy groups gives an isomorphism $\pi_2(S^2)\cong \pi_2(E)$ which is induced by the inclusion $i$.

Therefore $\pi_2(S^2)\cong \pi_2(E)\cong H_2(E;\mathbb{Z})\cong \mathbb{Z}$.

Let $\alpha,a,A$ be the generators of $H_2(E;\mathbb{Z}),\pi_2(E),\pi_2(S^2)$ respectively. Then $[i]=n\alpha$ for some $n\in \mathbb{Z}$. Under the isomorphism $H_2(E;\mathbb{Z})\cong \pi_2(E)$, the class $[i]$ maps to $\pm na$. Under the isomorphism $\pi_2(E)\cong \pi_2(S^2)$, $na$ maps to $\mp nA$.

My guess is that $[i]$ is a generator, but I'm not sure where to go from here. Probably I need to use the fact that $i$ induces the isomorphism $\pi_2(S^2)\cong \pi_2(E)$.

The map $i : S^2 \to E$ induces a map $i_* : \pi_2(S^2) \to \pi_2(E)$ given by $[f] \mapsto [i\circ f]$. The generator $[\operatorname{id}_{S^2}]$ of $\pi_2(S^2) \cong \mathbb{Z}$ is mapped to $[i] \in \pi_2(E)$; as $i_* : \pi_2(S^2) \to \pi_2(E)$ is an isomorphism, $[i]$ is a generator of $\pi_2(E) \cong \mathbb{Z}$. As $E$ is simply connected, the Hurewicz map $\pi_2(E) \to H_2(E)$ given by $[f] \mapsto f_*[S^2]$ is an isomorphism. Therefore, $i_*[S^2]$ is a generator of $H_2(E) \cong \mathbb{Z}$.
• Thank you for your answer. Does the inclusion $i$ also induce an isomorphism $H^2(E;\mathbb{Z})\to H^2(S^2;\mathbb{Z})$? – rpf Jun 10 '18 at 21:47