Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I made the following claim over at the Secret Blogging Seminar, and now I'm not sure it's true:

Let f: X --> Y and g: X --> Y be two maps betwen finite CW complexes. If f and g induce the same map on pi_k, for all k, then f and g are homotopic.

Was I telling the truth?

EDIT: Since I didn't say anything about basepoints, I probably should have said that f and g induce the same map

[S^k, X] --> [S^k, Y].

This will also deal better with the situation where X and Y are disconnected. I'd be interested in knowing a result like this either with pointed maps or nonpointed maps. (Although, of course, if you work with pointed maps you have to take X and Y connected, because [S^k, _] can't see anything beyond the number of components in that case.)

share|improve this question

4 Answers 4

up vote 19 down vote accepted

This is not true. Consider, for example, a degree 1 map from a torus S^1 \times S^1 to S^2 (concretely, realize the torus as a square with identifications, and then collapse the boundary of the square to a point). This map is trivial on all homotopy groups (since for any n>0, \pi_n is 0 for either the domain or the codomain), but it is not homotopically trivial because it is nonzero on H_2.

If you want to demand that the spaces be simply connected, you can get a counterexample by considering cohomology operations: the cup square, for example, gives a map from K(Z,n) to K(Z,2n) which is nontrivial, but for the same reason as the previous example it must be 0 on homotopy groups. This example is not finite-dimensional, but it's probably possible to find one that is--I just don't know how because I don't know how to show a map is trivial on homotopy groups if the spaces have infinitely many nontrivial homotopy groups whose values are unknown, which is the case for most finite-dimensional examples.

share|improve this answer
    
Thank you very much! –  David Speyer Oct 26 '09 at 20:53
3  
A further example from Allen Hatcher: The same thing works for the quotient map S^n x S^n ---> S^{2n} collapsing S^n v S^n to a point. This map is trivial on homotopy groups since \pi_i(S^n x S^n) = \pi_i(S^n) x \pi_i(S^n), so the inclusion of S^n v S^n into S^n x S^n is surjective on \pi_i. –  Eric Wofsey Oct 27 '09 at 17:53

Another interesting counterexample is given by so-called "phantom maps", which induce the zero map on all homotopy groups but are not nullhomotopic. Given an infinite CW-complex X which is a union ∪Xn of finite subcomplexes, Milnor described homotopy classes of maps out to Y where the phantom maps are given by a "lim1"-term.

For example, I believe that using this Brayton Gray used this to construct a map from CPinfinity to S3 that is nullhomotopic on CPn for all n.

share|improve this answer
    
It is worth pointing out that phantom maps are not uncommon. Indeed, in a compactly generated triangulated category an object which is not the target of any non-zero phantom map is pure-injective and the isomorphism classes of indecomposable pure-injective objects form a set. –  Greg Stevenson Oct 27 '09 at 2:20

For finite spectra, your question is precisely Freyd's generating hypothesis, which is open.

share|improve this answer

I think the proper Whitehead for maps says that if the cone of the map has trivial homotopy groups, then the map is a homotopy equivalence.

Edit: see also the discussion of Whitehead theorem in the comments.

share|improve this answer
    
Interesting, but I want a theorem whose conclusion is that f and g are homotopic. Thanks, though! –  David Speyer Oct 26 '09 at 21:17
    
Philosophically, Whitehead characterizes properties of one object, not two. You could get what you want by considering properties of a map А between maps f and g that is a candidate for homotopy, I think. –  Ilya Nikokoshev Oct 26 '09 at 21:26
3  
Ilya, this is not generally true unless the spaces in question are simply connected. –  Charles Rezk Oct 26 '09 at 22:28
4  
If you have a map f whose mapping cone is contractible, then excision in homology tells you that f induces an iso H_*X --> H_*Y in homology. The Whitehead theorem (or is it the Hurewicz theorem? I can never remember) tells you that if X and Y are simply connected CW complexes, then f is a homotopy equivalence. Counterexamples exist if the spaces aren't simply connected, however. –  Charles Rezk Oct 27 '09 at 0:25
2  
The version he's stating for simply-connected spaces is sometimes called the homology Whitehead theorem, which is a consequence of the ordinary Whitehead theorem and the Hurewicz theorem together. –  Tyler Lawson Oct 27 '09 at 0:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.