Whitehead for maps - MathOverflow

Whitehead for maps

2009-10-26T20:32:08Z

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.)

Answer by Eric Wofsey for Whitehead for maps

2009-10-26T20:51:48Z

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.

Answer by Ilya Nikokoshev for Whitehead for maps

2009-10-26T21:12:13Z

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.

Answer by Tyler Lawson for Whitehead for maps

2009-10-26T23:06:38Z

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 ∪X_n 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 CP^infinity to S³ that is nullhomotopic on CPⁿ for all n.

Answer by Josh Shadlen for Whitehead for maps

2009-10-27T04:44:21Z

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