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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have a surjective map, say $f$, between two spheres (of dimension $n+1 \geq 2$) such that it takes the closed upper hemisphere to itself and the closed lower hemisphere to itself. Now, I get a map $\hat{f}$ between $S^n$ by restricting $f$ to the equator.

Can someone help me find an explicit homotopy between the suspension of $\hat{f}$ and $f$

Using this I want to establish that the degree of $\hat{f}$ is the same as $f$, which I know to be true via another argument.

share|cite|improve this question

Let $S^+$ be be northen hemisphere of your $n$-sphere $S$. Then $f$ restricts to a map $f:(S^+,\partial S^+)\to(S^+,\partial S^+)$, which gives us a class in $\pi_n(S^+,\partial\ S^+)$. The long exact sequence for homotopy groups corresponding to the pair $(S^+,\partial S^+)$ shows that the boundary map $$\pi_n(S^+,\partial S^+)\to\pi_{n-1}(\partial S^+)$$ is an isomorphism.

Can you see how to use this fact (and the symmetric one for the southern hemisphere) to show what you want?

(Knowing how the long exact sequence for the homotopy groups of a pair is built out of the Barratt-Puppe sequence will help here; standard textbooks should help ---I particularily like the meticulousness of Switzer!--- as well as this nice post by our co-M.SEer Akhil)

share|cite|improve this answer
(Maybe one can build an exact sequence in homotopy for triples $(X;A,B)$ with $A$ and $B$ subspaces of $X$, and deduce what you want from some general statement) – Mariano Suárez-Alvarez Oct 10 '11 at 0:04

Your Answer


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.