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.

  • 2
    $\begingroup$ 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? $\endgroup$ Dec 6, 2019 at 1:20
  • 1
    $\begingroup$ This space is already simply connected hence is the universal cover of itself. Perhaps you intended the question about $\pi_2(S^2\vee S^1)$? $\endgroup$ Dec 6, 2019 at 1:28
  • $\begingroup$ Frankly speaking I am lost @EricWofsey I know that I can not use cohomology as I did not take it $\endgroup$
    – Secretly
    Dec 6, 2019 at 1:35
  • $\begingroup$ we are working from Hatcher@EricWofsey $\endgroup$
    – Secretly
    Dec 6, 2019 at 1:36
  • 2
    $\begingroup$ Well, this problem is solved (in a more general form) in the text of Hatcher in Example 4.26. $\endgroup$ Dec 6, 2019 at 1:38

2 Answers 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 $\pi_2(S^2 \vee S^2) \cong H_2(S^2 \vee S^2) \cong \mathbb{Z} \oplus \mathbb{Z}$.


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 $\pi_2(S^2\vee S^2) \cong \pi_2(S^2\times S^2)\cong \mathbb{Z}\oplus\mathbb{Z}.$

  • $\begingroup$ How do you get this exact sequence and why the operation between the two sphere is changed from $\wedge$ to $\vee$ to $\times.$ $\endgroup$
    – user591668
    Dec 6, 2019 at 2:23
  • $\begingroup$ @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. $\endgroup$ Dec 6, 2019 at 2:46

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