Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f : S^n \rightarrow S^n$ be a continuous map. One can define the degree of $f$ as the integer $k$ such that $f_* : H_n(S^n) \rightarrow H_n(S^n)$ is multiplication by $k$ on $H_n(S^n) \cong \mathbb Z$. However, it also makes sense so define the degree as the integer $k'$ such that $f^*$ is multiplication by $k'$ on $H^n(S^n)$. Must these two integers agree?

share|improve this question

1 Answer 1

up vote 5 down vote accepted

Yes.

From the universal coefficient theorem for cohomology, we have a natural isomorphism $H^n(S^n)\to\hom(H_n(S^n),\mathbb Z)$. Now write what naturality means for a map $f:S^n\to S^n$, and conclude what you want.

share|improve this answer
    
This also tells you what happens for the degree of a map $f:M\to M$ with $M$ a closed orientable manifold, more generally, btw. –  Mariano Suárez-Alvarez Jan 13 '13 at 4:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.