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Show that $$f_* \colon H_4(S^4) \to H_4(S^2 \times S^2)$$ is the zero map for any $f\colon S^4 \to S^2 \times S^2$.

We are working with integral coefficients. I tried applying the naturality of Künneth Theorem, obtaining the following commutative diagram (Tor vanishes)

enter image description here

But I'm unsure how does the map "?" look, my idea was to use the map $f$ and projection on the two factors, but in this case I'm not able to prove that it is the zero map here. I don't have any other idea on how to compute $f_*$

Luigi M
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    It's easier to see the corresponding statement about cohomology: the point is that a generator of $H^4(S^2 \times S^2)$ is given by the cup product of two classes in $H^2$, but $H^2(S^4) = 0$. – Qiaochu Yuan May 13 '15 at 21:28
  • I'll try thinking about that, I don't have a good feeling with the cup product, (it was explained to me 3 days ago). Then passing to homology should be easy? – Luigi M May 13 '15 at 21:36
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    Since $H^2(S^4) = 0$ and cup product commutes with pullback, you'll get that the pullback of a generator of $H^4(S^2 \times S^2)$ by $f$ must be 0. But then by definition of the pullback, $f_* = 0$. – William Stagner May 13 '15 at 21:41
  • I'm not sure, but I think you could write $f$ as a composition $S^4 \to S^4 \times S^4 \to S^2 \times S^2$ and then use the naturality of the Künneth isomorphism to show that the induced map $H_4(S^4 \times S^4) \to H_4(S^2 \times S^2)$ is zero. – Math536 May 14 '15 at 03:54
  • @Math536 this seems very similar to what I've in mind, but I've problems in showing that the induced map from $H_4(S^4\times S^4) \to H_4(S^2\times S^2)$ is zero. – Luigi M May 14 '15 at 07:37
  • @QiaochuYuan the fact that a generator for $H_4(S^2\times S^2)$ is a consequence of the kunneth theorem for cohomology (b/c S^2 is a finite cw) or comes from a more general statement-setting? (Hope the question is meaningful, I'm getting used of these machinery now) – Luigi M May 14 '15 at 09:35
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    @Luigi: yes, it follows from the Kunneth theorem. – Qiaochu Yuan May 14 '15 at 16:33
  • @Luigi:You can express the volume element of $S^2\times S^2$ as a wedge product of the volume elements of the two spheres. These pull back to closed 2 forms in $S^4$ that must be exact since $H^2(S^4) = 0$. This is the de Rham equivalent of the cup product argument, – Joe S May 14 '15 at 21:36

1 Answers1

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Künneth for products says $H^4(S^2 \times S^2)$ is generated by $s=\pi_X^* x \smile \pi_Y^* y$ where the pis are the projections and $x,y$ generators of the top cohomology of $X$,$Y$ resp. Since cup product is natural, in other words cohomology is a functor to graded rings, we get $$f^*(s) = f^*(\pi_X^* x \smile \pi_Y^* y) = f^*(\pi_X^* x) \smile f^*(\pi_Y^* y) =0 \smile 0 =0$$

since $H^2(S^4)=0$. Now what does that imply for the induced map on homology?

you could also use that $\pi_4(S^2\times S^2)=\pi_2(S^2)^2=$ only torsion (by Serre). For example by post composing $f$ with a map to $S^4$ which is not trivial on top homology for sure.

Daniel Valenzuela
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