Why every map $f : S^n \to T^n (n>1)$ has topological degree zero?

Question

Why every map $f : S^n \to T^n (n>1)$ has topological degree zero?

I don't know anything about covering spaces, and has been told to me that this assertion comes from this theory!

I do appreciate any help.

Answer 1

Since $S^n$ is simply-connected for $n>1$, any such map $f$ factors through $\mathbb{R}^n$, which is the universal cover of $T^n$ and has $n$-th degree trivial homology since its contractible. It follows that the $n$-th degree induced map of $f$ on homology is trivial by functoriality of homology. The key point here is the Lifting Theorem.

Answer 2

One can do this without covering spaces using the cup product in cohomology. Namely $H^n(T^n,\Bbb Z)=\Bbb Z\langle e_1\cup e_2\dots \cup e_n\rangle$ (this is supposed to denote integer multiples of $ e_1\cup e_2\dots \cup e_n$), where $e_i$ are the generators of $H^1(T^n,\Bbb Z)$. Now if $f:S^n \to T^n$ has nonzero topological degree then $f^*( e_1\cup e_2\dots \cup e_n)$ is nonzero. But by naturality of the cup product $f^*( e_1\cup e_2\dots \cup e_n)=f^*(e_1)\cup f^*(e_2) \cup \dots f^*(e_n)$, which is zero since $H^1(S^n,\Bbb Z)=0$.