# Homotopy class of map from torus to sphere?

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 $$\mathbb{Z}$$.

• 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. – jgon Nov 5 '18 at 22:32

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.
• 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 $T^n\rightarrow S^n$. – maplemaple Nov 5 '18 at 22:38
• @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. – Qiaochu Yuan Nov 5 '18 at 22:43