Compute $\pi_{2}(S^2 \vee S^2).$

Question

Compute $\pi_{2}(S^2 \vee S^2).$

Hint: Use universal covering thm. and use Van Kampen to show it is simply connected.

Still I am unable to solve it, could anyone give me more detailed hint and the general idea of the solution.

Answer 1

Following homotopy excision theorem and using an exact sequence of a pair $(S^2\times S^2, S^2\vee S^2)$ you can write down an exact sequence $$ 0 \to \pi_3(S^2\wedge S^2)\to \pi_2(S^2\vee S^2) \to \pi_2(S^2\times S^2) \to 0 $$ Since $S^2\wedge S^2 \simeq S^4$ you have an isomorphism $\pi_2(S^2\vee S^2) \cong \pi_2(S^2\times S^2)\cong \mathbb{Z}\oplus\mathbb{Z}.$

Answer 2

Here's another quick way, using Hurewicz.

$\pi_1(S^2 \vee S^2) \cong 0$ by van Kampen. Then the Hurewicz theorem asserts that $\pi_2(S^2 \vee S^2) \cong H_2(S^2 \vee S^2) \cong \mathbb{Z} \oplus \mathbb{Z}$.