In algebraic topology, we use homotopy group $\pi_n(M)$ to classify the homotopy class of continuous map $f:S^n\rightarrow M$.

My question:

  1. What's the mathematical object to classify the homotopy class $f: N \rightarrow M$ for general manifold $N$ and $M$? Can we reduce this question to $\pi_n (N)$ and $\pi_n (M)$?

  2. If question 1 is too hard. In specific, how to classify homotopy class of $f: T^2 \rightarrow S^2$ and $f: T^n\rightarrow S^n$ for general $n$? Thanks to Qiaochu Yuan, the answer to homotopy group of $T^n$ to $S^n$ seems to $\mathbb{Z}$.

  • $\begingroup$ The mathematical object is just the set of homotopy classes of maps from $N$ to $M$. You could call it $\operatorname{Hom}(N,M)$ in the homotopy category if you wanted. $\endgroup$ – jgon Nov 5 '18 at 22:32

In general it is extremely hard to calculate the set of homotopy classes of maps between two spaces, and the problem does not reduce to understanding their homotopy groups. This problem has as special cases questions like 1) how to compute an arbitrary cohomology group of an arbitrary space, 2) how to compute an arbitrary homotopy group of an arbitrary space, 3) how to classify vector bundles on an arbitrary space, etc.

For the very special case of maps from a closed orientable $n$-manifold (such as the torus $T^n$) to a sphere $S^n$ of the same dimension we are very lucky: by the Hopf theorem such maps are classified by their degree, so there are a $\mathbb{Z}$'s worth of them.

  • $\begingroup$ Thank you so much. Where can I find the proof of my 2nd question? i.e. Homotopic class is $\mathbb{Z}$. And I want the specific construction of map from $T^n\rightarrow S^n$. $\endgroup$ – maplemaple Nov 5 '18 at 22:38
  • $\begingroup$ @maplemaple: I don't know how Hopf proved it. I have in mind an annoying proof using obstruction theory. To construct a map from a closed oriented $n$-manifold $M$ to $S^n$, pick a point in $M$ and a small open neighborhood of that point looking like $\mathbb{R}^n$. Then collapse the rest of $M$ to a point. This gives a map $M \to S^n$ of degree $1$, and then composition with maps $S^n \to S^n$ of other degrees gives maps $M \to S^n$ of other degrees. $\endgroup$ – Qiaochu Yuan Nov 5 '18 at 22:43
  • $\begingroup$ What happens with the Hopf degree theorem if we're just in the continuous, PL, etc. category? The proof I'm familiar with uses some results about transversality that don't apply outside the smooth case. $\endgroup$ – anomaly Nov 5 '18 at 22:52
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    $\begingroup$ @anomaly: for maps between manifolds it should all be the same, in that every homotopy class of continuous maps between smooth or PL manifolds should have a smooth or PL representative. The proof I have in mind doesn't need this. $\endgroup$ – Qiaochu Yuan Nov 5 '18 at 23:15
  • $\begingroup$ @QiaochuYuan: Cool, thanks. $\endgroup$ – anomaly Nov 6 '18 at 2:29

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