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I'm asked to present a continuous function $\alpha \colon S^3 \rightarrow S^2 \vee S^2$ s.t. it is not null homotopic but taken both projections $pr \colon S^2 \vee S^2 \rightarrow S^2$ the composition $pr \circ \alpha$ is null homotopic.

I'm given the hint to use tha fact that $S^2 \times S^2$ is not homotopic equivalent to $S^2 \vee S^2 \vee S^4$ (they cohomology rings are not isomorphic).

Any help or advice is well accepted: thanks in advance.

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2 Answers 2

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Consider the attaching map $S^3 \to S^2 \vee S^2$ of the $4$-cell in the standard CW-structure of $S^2 \times S^2$. If this is nullhmotopic, $S^2 \times S^2$ would be homotopic to $S^2 \vee S^2 \vee S^4$ as homotopic attaching map implies homotopy equivalent spaces.

But cup square $\alpha \smile \alpha$ is nontrivial in $H^*(S^2 \times S^2)$ where $\alpha$ is generator of $H^2(S^2)$, whereas it's trivial in $H^*(S^2 \vee S^2 \vee S^4) \cong H^*(S^2) \times H^*(S^2) \times H^*(S^4)$, contradiction.

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  • $\begingroup$ @MikeMiller Could you elaborate? $\endgroup$
    – Kyle
    Commented Dec 14, 2015 at 14:54
  • $\begingroup$ @Kyle: Ah, that was complete nonsense. Your original comment's suggestion - that we try working with the Hopf map $S^3 \to S^2 \to S^2 \vee S^2$ - does work; the cell structure you get (as in this answer) is $\Bbb{CP}^2 \vee S^2$, which has the wrong cup product structure to be trivial or to be what's in this answer. $\endgroup$
    – user98602
    Commented Dec 14, 2015 at 14:59
  • $\begingroup$ @MikeMiller: Cool. I think we'd need to use $\mathbb{CP}^2 \vee \mathbb{CP}^2$ or something to get both projections to $S^2$ to be nontrivial, using a composition $S^3 \to S^3 \vee S^3 \to S^2 \vee S^2$. $\endgroup$
    – Kyle
    Commented Dec 14, 2015 at 15:10
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From the LES in homotopy groups for the pair $(S^2 \times S^2, S^2 \vee S^2)$, we have an exact sequence $$\pi_4(S^2 \vee S^2) \to \pi_4(S^2 \times S^2) \to \pi_4(S^2 \times S^2, S^2 \vee S^2)\to \pi_3(S^2 \vee S^2) \to \pi_3(S^2 \times S^2)$$

Since we have $\pi_k(S^2 \times S^2)\cong \pi_k(S^2 ) \times \pi_k( S^2) \subset \pi_k(S^2 \vee S^2)$, the maps $\pi_k(S^2 \vee S^2) \to \pi_k(S^2 \times S^2)$ are surjective. It follows that $\pi_4(S^2 \times S^2)\to \pi_4(S^2 \times S^2, S^2 \vee S^2)$ is the zero map, giving us a short exact sequence $$0 \to \pi_4(S^2 \times S^2, S^2 \vee S^2)\to \pi_3(S^2 \vee S^2) \to \pi_3(S^2 \times S^2) \to 0.$$ We want an element $\alpha \in \pi_3(S^2 \vee S^2)$ such that its composition with each projection $S^2 \vee S^2 \to S^2$ is nullhomotopic. These projections factor through $S^2 \times S^2$, so $\alpha$ lies in the kernel of the composition $\pi_3(S^2 \vee S^2) \to \pi_3(S^2 \times S^2) \overset{\sim}{\to} \pi_3(S^2) \times \pi_3(S^2)$. Thus it suffices to show that $\pi_4(S^2 \times S^2, S^2 \vee S^2)$ is nontrivial, since any nonzero element in its image in $\pi_3(S^2 \vee S^2)$ will work for us.

Here's where the hint comes in: Recall that $S^2 \times S^2$ has a cell structure with one $0$-cell, two $2$-cells, and one $4$-cell. The 3-skeleton is simply $S^2 \vee S^2$, so the boundary of the 4-cell is mapped into $S^2 \vee S^2$. Since $S^2 \vee S^2 \vee S^4$ is not homotopy equivalent to $S^2 \times S^2$, the 4-cell's attaching map is nontrivial, giving us the desired element of $\pi_3(S^2 \vee S^2)$.

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    $\begingroup$ I got the main idea. Unluckily I don't know homotopy theory so I cannot use your proof (but from what I understand it should be correct). I made an ad hoc proof that the attaching map $\alpha$ you propose has the property that $pr \circ \alpha$ is null homotopic: consider the characteristic map $Q \colon D^4 \rightarrow S^2 \times S^2$ associated to $\alpha$, as you said the map $pr$ factorizes through $S^2 \times S^2$ so we have the equality $pr \circ \alpha= \pi \circ Q \circ i$ where $\pi\colon S^2 \times S^2 \rightarrow S^2$ is a projection (continue) $\endgroup$
    – N.B.
    Commented May 28, 2015 at 15:44
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    $\begingroup$ and $i \colon S^3 \hookrightarrow D^4$ the inclusion. But $D^4$ is contractible and it's easy to see that $i$ must be null homotopic leading to the fact that the whole composition $\pi \circ Q \circ i$ is null homotopic as desired. $\endgroup$
    – N.B.
    Commented May 28, 2015 at 15:44
  • $\begingroup$ @N.B.: Nice, that's a better argument for showing that $\operatorname{pr}\circ \alpha$ is nullhomotopic. The homotopy-theoretic argument was just me thinking out loud to find a candidate $\alpha$. As you've shown, you can use more elementary techniques to prove that a given $\alpha$ works. $\endgroup$
    – Kyle
    Commented May 28, 2015 at 15:56

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