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

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.

• Where is this problem from? What is the "universal covering thm." (which does not sound like it would be relevant to this problem...)? What tools do you have available? Dec 6, 2019 at 1:20
• This space is already simply connected hence is the universal cover of itself. Perhaps you intended the question about $\pi_2(S^2\vee S^1)$? Dec 6, 2019 at 1:28
• Frankly speaking I am lost @EricWofsey I know that I can not use cohomology as I did not take it Dec 6, 2019 at 1:35
• we are working from Hatcher@EricWofsey Dec 6, 2019 at 1:36
• Well, this problem is solved (in a more general form) in the text of Hatcher in Example 4.26. Dec 6, 2019 at 1:38

$$\pi_1(S^2 \vee S^2) \cong 0$$ by van Kampen. Then the Hurewicz theorem asserts that $$\pi_2(S^2 \vee S^2) \cong H_2(S^2 \vee S^2) \cong \mathbb{Z} \oplus \mathbb{Z}$$.
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 $$\pi_2(S^2\vee S^2) \cong \pi_2(S^2\times S^2)\cong \mathbb{Z}\oplus\mathbb{Z}.$$
• How do you get this exact sequence and why the operation between the two sphere is changed from $\wedge$ to $\vee$ to $\times.$
• @Mathstupid From homotopy excision $\pi_3(S^2\times S^2, S^2\vee S^2)\cong \pi_3(S^2\times S^2/S^2\vee S^2) = \pi_3(S^2\wedge S^2).$ The exact sequence is just a part of the exact sequence of a pair $(S^2\times S^2, S^2\vee S^2),$ as I've written above. Dec 6, 2019 at 2:46