Degree in homology and cohomology

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?

-

1 Answer

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.

-
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