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Am I correct in saying that a space $X$ has trivial $n$th homotopy group if and only if every continuous function from the $n$-sphere to $X$ can be extended to a continuous function from the $n+1$-ball?

Is there a similar simple interpretation of having trivial $n$-th homology group? I am especially interested in understanding what it intuitively means that for $k>n$, the $k$-th homology group of the $n$-sphere is trivial, which contrasts with homotopy groups.

For instance, the 3rd homology group of the 2-sphere is trivial. However the 3rd homotopy group is not, which is witnessed by the Hopf fibration, which is a continuous function $f$ from the 3-sphere to the 2-sphere that cannot be extended to the 4-ball. From it, one can define a 3-cycle of the 2-sphere. As the 3rd homology group is trivial, that cycle is a boundary. What does it mean, concretely? Maybe that there is a space, obtained by gluing together a finite number of 4-simplices, whose boundary is homeomorphic to the 3-sphere, on which an extension of $f$ can be defined?

mathieu
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    You can find relevant discussion for instance in Hatcher's book. The key is to replace maps of spheres/balls by maps of pseudomanifolds/pseudomanifolds with boundary. – Moishe Kohan Nov 26 '20 at 16:12
  • Thanks! Ok, so from the discussion in the Hatcher, a 2-cycle corresponds to a function defined on a closed oriented surface $S$, and it is a boundary if that function can be extended to a manifold whose boundary is the surface $S$. However, such a correspondance with manifolds no more holds for higher-dimensional cycles. So maybe there is no such simple understanding of having trivial homology groups. – mathieu Nov 27 '20 at 08:00
  • See my answer [here](https://math.stackexchange.com/questions/40149/intuition-of-the-meaning-of-homology-groups), and other answers too might be helpful. – Moishe Kohan Nov 27 '20 at 10:50
  • This is indeed very helpful and fully answers my question. Many thanks! – mathieu Nov 30 '20 at 08:29

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