# Hopotopy type question

1. Apr 14, 2007

### quasar987

1. The problem statement, all variables and given/known data
Is it true that if X x Y has the same homotopy type as X' x Y', then X has the same homotopy type as X' or Y' and Y has the same homotopy type as Y' or X' (the one not already in use by X).

3. The attempt at a solution

I've tried finding a counter-exemple, but no luck. I've also established that the (natural) procedure to show that $X\simeq X', \ \ Y\simeq Y' \ \Rightarrow X\times Y \simeq X'\times Y'$ does not work in the other direction (i.e. we can't go backward from explicit homotopies btw X x Y and X' x Y' to explicit homotopies btw X and Y and X' and Y')

2. Apr 15, 2007

### matt grime

It is easy to think of counter examples if you're simple minded. I.e. don't try to think of producting together complicated spaces for a counter example. You can get one with just products of circles involved.

It's asking if, whenever you write a space as a product of two spaces, then there is only one possible choice of pair of homotopy type for the two spaces. This can't be true, we feel, and sure enough if we just pick *any* space that is the product of two non-contractible spaces we're going to have a good chance of getting a counter example.

So all I've asked for is a space

Z=XxY

Can you see what to do now (given I don't apparently have an X' and a Y' lying around?)

3. Apr 15, 2007

### matt grime

Of course a more interesting question* is:

suppse that XxY and X'xY are homotopic. Does the cancellation property hold? I.e. is X homotopic to X'

* you original question is almost never going to be true. It's like me asking: if N=p*q and N=r*s does it follow thaht p=r (or p=s). That's only true for primes or products of two primes.

Last edited: Apr 15, 2007
4. Apr 15, 2007

### mathwonk

the original question fails even when exactly one of the spaces is (homotopic to) a point.

i don't have a candidate for the second question.

i might be able to think up a situation where the product spaces both have the same homology (say trivial) but not the original ones.

using kunneth, one could try for some torsion groups that kill each other off under tensoring.

no wait, what about taking a product of an infinite number of circles, first with a point, then with a circle? here you even get homeomorphism of the products, but the factors X,X' are not even homotopy equivalent.

5. Apr 15, 2007

### quasar987

6. Apr 15, 2007

### StatusX

One trivial counterexample comes from taking one of the spaces to be a point, in which case the product is homeomorphic to the other factor.

7. Apr 15, 2007

### quasar987

You're saying set X={p} and let Y be anything. Then, X x Y ~ Y. But this is does not meet the hypothesis. The hypothesis is that X x Y has the same homotopy type as some product X' x Y'.

8. Apr 15, 2007

### StatusX

Just think a little harder.

9. Apr 15, 2007

### quasar987

Stop torturing me, I've explored all logical extension and implications available to all 4 of my neural links. Plus, the exam is in 16 hours.

Mercy!

10. Apr 15, 2007

### StatusX

Take Y itself to be a product, such that Y is not homotopy equivalent to either of its factors (eg, S^1 x S^1).

Get some sleep before the test. It's much more important for you to be able to think up simple things like this than anything you might memorize while studying.

11. Apr 15, 2007

### quasar987

Don't worry, it's only 17h35 here and the exam is tomorrow morning at 10h30 so I have time to do both.

And thank you. Maybe I would have found it after thinking for 10 years.

12. Apr 15, 2007

### matt grime

What that S^1xS^1 and {pt}x(S^1xS^1)

are a counter example? I'd hope not. Just think about tori (i.e. T^n is S^1xS^1x..xS^1 n-1 times).

Just forget homotopy, since that is misleading. Are you saying that you can't think of reasonalbe counter examples to the assertion: however I write a space as the product of two spaces, then I get the same answer (up to some level of equivalence)?

@mathwonk: allowing infinitely many terms would always produce a counter example. It's just Krull-Schmidt. I have an interesting (open, hard) problem on K-S for triangulated categories. Perhaps I'll ask you in August when I shouold be in Athens.

Last edited: Apr 15, 2007