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Let as have a continuous map from a topological space X to an n-sphere, realized as a simplicial map between simplicial complexes (for example), i.e. represented by a finite set of data. Is there an algorithm for determining, whether or not the map is homotopic to a constant map?

I know that for maps from a circle to a 2-complex, this is impossible (due to the word problem). But is it possible for maps "to the sphere"? If the dimension of X is less then 2n-2, a group structure may be defined on the cohomotopy space $\pi^n(X)$ and this group is always commutative, so, the obstacle may be not so hard as in case of a "fundamental group" element.

Sorry if this is not too exact, I hope it makes at least some sense :-).

Peter

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  • $\begingroup$ As I mentioned in the comments to your previous thread, when the target is the circle your question can be interpreted as whether or not an element of $H^1(X;\mathbb Z)$ is trivial, which is a computable problem. When the target space is $S^n$ for $n \geq 2$ there is a finite sequence of cohomological obstructions to a null-homotopy. I suggest looking at Whitehead's book "Elements of Homotopy theory" section V.5 on "Obstruction Theory". This isn't quite in the form you could implement (in full generality) on a computer since at some point you run out of knowledge of homotopy grp of spheres. $\endgroup$ Commented Mar 8, 2011 at 16:08
  • $\begingroup$ @Ryan: I will read that paper, thank you very much for all the comments. $\endgroup$ Commented Mar 8, 2011 at 17:14
  • $\begingroup$ It's also worth noting that the condition $\dim(X)\leq 2n-2$ ensures that $[X,S^n]=[X,\Omega S^{n+1}]=[\Sigma X,S^{n+1}]$ and thus (by induction and passage to the limit) that $[X,S^n]$ is the same as the stable group $[X,QS^n]=[\Sigma^\infty X,\Sigma^\infty S^n]$. Thus, you are really doing stable homotopy theory, which makes the obstructions significantly easier. In fact, it is an old theorem of Brown that these groups are recursively computable. It seems to be folklore that the proof leads to completely impractical algorithms, but I have never tried to check that. $\endgroup$ Commented Mar 8, 2011 at 23:50

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