Re: Cohomology, homology, cup product, spheres, question

James wrote:
My question is to show that any map S^4 ---> S^2 x S^2 must induce
the zero homomorphism on H_4.

Using cup product arguments, I have shown that any map S^4 ----> S^2
x S^2 induces the zero homomorphism on H^4. I want to somehow now
use the naturality of the Universal coefficients theorem, but I just
can't get it.

Of course all Ext groups vanish in the Universal Coefficients
theorem. The question is : In this case, if I showed that the zero
homomorphism is induced on H^4, does that mean we get a zero
homomorphism on H_4? Why?

Thank you for your time and help,

James

Let f : S^4 ---> S^2 x S^2.

From UCT, you have the natural short exact sequences

Ext(H_3(S^2 x S^2), Z) >--> H^4(S^2 x S^2) -->> Hom(H_4(S^2 x S^2), Z)
| | |
f% | f* | . o f_*|
| | |
V V V
Ext(H_3(S^4), Z) >---> H^4(S^4) --->> Hom(S^4 , Z)

where the vertical homomorphisms are induced from the map f (the first
and last are induced from f_* by contravariance of Ext and Hom, wrt the
first variable, the middle is f*, the homomorphism on cohomology. As
you noticed, the groups on the left are both zero, so this diagram is:

H^4(S^2 x S^2) >-->> Hom(H_4(S^2 x S^2), Z)
| |
f* | . o f_*|
| |
V V
H^4(S^4) >--->> Hom(S^4 , Z)

Since the homomorphism f* is zero (as you noticed, by virtue of
H^4(S^2 x S^2) being generated by the cup product of 2-dimensional
classes, the homomorphism . o f_* (which is, as the notation suggests,
composition of f_* with an element of Hom(H_2(S^2 x S^2),Z)) is
also zero. Now, let f_* take the generator [S^4] to the multiple
k[S^2 x S^2] of H_4(S^2 x S^2). Then, taking a nonzero homomorphism
h of H_4(S^2 x S^2) to Z, it would take the generator [S^2 x S^2]
to some nonzero integer m. Composing h with f_*, we find

[S^4] |----> k [S^2 x S^2] |----> km

Since m is assumed nonzero, k*m is zero (as required) only if k = 0.

Done.

Dale.
.

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