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Hi all. If n>m and I have a continuous map from an n-sphere to an m-sphere, is there a simple way to see whether the map is homotopic to a constant? (For example, is it if and only if the map is not onto? -- this would be probably too simple). For n=m, the answer is "nonzero degre" -- is there a reasanable generalisation of "degree" to different dimensions? Can you recommend me some literature about it? Thanks, Peter

flag	asked Dec 30 2010 at 11:48 peter franek 317●5

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As you suggest, if the map misses a single point then its image lies in a disc inside the sphere, and contracting the disc to a point gives a null-homotopy. The converse is not true - even if n=m, there are degree zero maps that are surjective. This is a very difficult and interesting problem. In general, calculating the homotopy groups of spheres is one of the central problems of algebraic topology. – Jeffrey Giansiracusa Dec 30 2010 at 12:24

Hi Jeffrey. Thanks for your comment. I know that homotopies of spheres are not known in general, but I thought that there might be some tool for recognizing whether the map is homotopically trivial. – peter franek Dec 30 2010 at 12:40

For $n<m$ there are continuous onto maps $S^n \to S^m$, yet all such maps are null-homotopic. What would you consider a reasonable answer in this case, Peter? i.e. would it be acceptable to just say all maps are null homotopic so you don't need to know anything about a particular map $S^n \to S^m$? The problem with that is you've accepted computing the homotopy groups of spheres as a prerequisite to your problem, if you go that direction. So there's nothing simple like degree. – Ryan Budney Dec 30 2010 at 17:48

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There is something similar to degree theory but it rarely helps you decide whether or not a given map $S^n \to S^m$ is null, see: en.wikipedia.org/wiki/… – Ryan Budney Dec 30 2010 at 18:02

My problem comes from informatics. I want to create a computer algorithm that decides, whether f=0 has solution. The function f is assumed to be "nice", from an n-cube (product of n intervals) to Rm, and we can over-approximate its image on small cubes arbitrary precise. I'm interrested in "robust" solution, i.e. such that f+g=0 has still solution for g small. I proved that f=0 has a robust solution iff the map f/|f| from the boundary of a neighborhood of a component of {f=0} to the sphere is homotopically nontrivial. Should I give up any hope, for n>m, to find such an algorithm?(probably yes) – peter franek Jan 27 2011 at 23:26

Dev Sinha · Answer 1 · 2010-12-31 04:26:40Z UTC

The set of maps $S^n \to S^m$ is better known as $\pi_n(S^m)$. The one other (than n=m) relatively easy case is when we tensor by the rational numbers (and thus answer the question "when is f homotopic to g after adding each to itself some number of times?). The answer in this case is they are always homotopic if m is odd or if m is even and $n \neq 2m - 1$. If $m$ is even and $n = 2m-1$ then $f$ and $g$ are homotopic in this weak sense if and only if they both have non-trivial (or both have trivial) classical Hopf invariant.

Jeff Strom · Answer 2 · 2011-01-01 06:08:59Z UTC

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Every map from a sphere to a finite connected CW complex is homotopic to a surjective one. Collapse one half of the sphere to a line segment, use your original map on the sphere created by the collapse, and map the segment surjectively onto the complex (which can be done by the Hahn-Mazurkowicz Theorem).

edited Jan 1 2011 at 6:08

answered Dec 31 2010 at 19:59

Jeff Strom
3,504●1●6●19

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finite connected CW-complex – Ryan Budney Dec 31 2010 at 20:59

thanks -- fixed now. – Jeff Strom Jan 1 2011 at 6:09

homotopy of sphere maps

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