Can You Prove $\pi_2(\Bbb CP^\infty) \approx \Bbb Z$?

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Here is the last POTW for 2014!

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Problem. Show that $\pi_2(\Bbb CP^\infty) \approx \Bbb Z$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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No one answered this week's problem. You can find my solution below.

Spoiler

There is a fiber bundle $S^1 \to S^\infty \to CP^\infty$, which induces a homotopy exact sequence

\(\displaystyle \pi_2(S^\infty) \to \pi_2(CP^\infty) \to \pi_1(S^1)\to \pi_1(S^\infty)\)

Since $\pi_2(S^\infty) = \pi_1(S^\infty) = 0$ and $\pi_1(S^1) \approx \Bbb Z$, the map $\pi_2(CP^\infty) \to \pi_1(S^1)$ gives an isomorphism $\pi_2(CP^\infty) \approx \Bbb Z$.