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Can you construct a map $$F: S^{1}\times\cdots\times S^{1}(n~\text{copies of}~S^{1})\rightarrow S^{n}$$ of nonzero degree?

Mikasa
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shrinklemma
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1 Answers1

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Consider a small standard closed ball $B$ in your torus $T$, and let $F$ be the complement in $T$ of the interior of $B$. Contracting $F$ to a point gives us a map $T\to T/F\cong S^n$.

What is its degree?

Mariano Suárez-Álvarez
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  • You mean contracting $F$ to a point? Then almost every point of $S^{n}$ has preimage of a point in $B$, and this map is a local homeomorphism near this point, so the degree is $\pm 1$. Am I right? – shrinklemma Mar 23 '13 at 07:20
  • So this method can actually show that there exists maps of any degree, right? You just need to find disjoint balls in the torus. And the domain can be any n-dimensional manifold? – shrinklemma Mar 23 '13 at 07:22
  • Since there exist maps $S^n\to S^n$ of any degree, composing them with mine give you maps $T\to S^n$ of any degree. – Mariano Suárez-Álvarez Mar 23 '13 at 07:24
  • You are right. Thanks a lot. – shrinklemma Mar 23 '13 at 07:27
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    +1 So this seems to work for any compact $n$-dimensional orientable manifold. Beautifully simple: thanks for sharing ! – Georges Elencwajg Mar 23 '13 at 08:20