2
$\begingroup$

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 $\pi_2(S^2)\cong \pi_2(E)$.

$\endgroup$
3
$\begingroup$

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 $H_2(E) \cong \mathbb{Z}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your answer. Does the inclusion $i$ also induce an isomorphism $H^2(E;\mathbb{Z})\to H^2(S^2;\mathbb{Z})$? $\endgroup$ – rpf Jun 10 '18 at 21:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.