I've been working on the exercise "Find $\pi_2 (S^2 / X)$ where $X$ is the image of $S^1 \vee S^1$ under some embedding into $S^2$".

First I tried to find some helpful fibrations/cofibrations as this is kind of what we did in the lecture last week. It wasn't very successful. Then I tried a more direct approach, i.e. drawing a picture of the situation.

Now: if I embedd $X$ into $S^2$ it will somehow distinguish $X$ into $3$ discs. Now taking the quotient, it seems pretty clear to me that $S^2 / X \cong S^2 \vee S^2 \vee S^2$. Of that I know $\pi_2$ to be $Z^3$.

Is my intuition correct? And is there an easy way to see it formally, e.g. using standard homeomorphisms for spheres such as smash or suspension?

  • $\begingroup$ Yes your intuition is correct. $\endgroup$ – Cheerful Parsnip Dec 6 '13 at 15:03

It's not that hard to prove that $S^2/X$ is homeomorphic to $S^2 \vee S^2 \vee S^2$ directly. If you can draw the picture with the three disks then you know where to send every point (think about the presentation of $S^2$ as $D^2 / \partial D^2$). It only remains to check that this map is a homeomorphism which you should be able to do.

For a much less elementary approach that also proves much less, one could argue as follows. $S^2/X$ is obviousy simply connected. So if we can compute $H_2(S^2/X)$, Hurewicz isomorphism will give us $\pi_2(S^2/X)$. But $H_2(S^2/X)$ is nothing else than the relative homology $H_2(S^2, X)$. From the long exact sequence for the pair $(S^2, X)$ $$ 0 = H_2(X) \to H_2(S^2) \to H_2(S^2, X) \to H_1(X) \to H_1(S^2) = 0 $$ we immediately get that $H_2(S^2/X) \cong H_2(S^2) \oplus H_1(S^1 \vee S^1)$. Note that the reason the sequence splits as well as the reason why I'm not writing coefficients for the homology is that there is no torsion in spheres in these low dimensions.


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