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For every $k \in {\mathbb Z}$ construct a continuous map $f: S^n \to S^n$ with $\deg(f) = k$.

How do I show that for any $n>0$ and any $m$ any integer, there exists a map $f: S^n\to S^n$ of degree $m$?

I am trying a constructive proof(constructing such a map) but I still did not find a map whose degree is not $1$ or $-1$. I am viewing this question from a singular homology point of view.

Thanks.

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Herband
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  • @Jason, yes. It is an exact duplicate and nicely stated there. Can you help with this question? – Herband Mar 01 '12 at 21:02
  • Can you find a nice degree $m$ map $f_{m,1}$ from $S^1$ to $S^1$?

    If so, try moving up to $S^2$. You can start with with $f_{m,1}$ as a map on the equator of $S^2$ and extend it to a nice map $f_{m,2}: S^2 \rightarrow S^2$. The same idea works in general to extend a degree $m$ map $f_{m,n}:S^{n} \rightarrow S^{n}$ to a degree $m$ map $f_{m,n+1}:S^{n+1} \rightarrow S^{n+1}$.

     – Jonas Kibelbek Mar 01 '12 at 21:03
  • For the case $n=1$, it is a lemma in Homotopy theory By Sze-Tsen Hu, page 38, http://books.google.it/books?id=iVhMPU0X2G4C&pg=PA37&lpg=PA37&dq=homotopy+classes+of+maps+from+-spheres+to+-spheres+are+classified+by+their+degree&source=bl&ots=IxpX3lCr7k&sig=RI9zQkI6wDAVCI0F_zVIVE8EzIk&hl=en&sa=X&ei=MttPT9mMMMazhAf9hrTQCw&redir_esc=y#v=onepage&q=homotopy%20classes%20of%20maps%20from%20-spheres%20to%20-spheres%20are%20classified%20by%20their%20degree&f=false – Herband Mar 01 '12 at 21:04
  • @Jonas, thank you. Can you show me the first extension? – Herband Mar 01 '12 at 21:06
  • Every horizontal slice of the sphere $S^2$ is just a circle $S^1$ (except for the north and south poles). You can use the map $f_{m,1}$ on each of these circles (and use the identity map on the poles) so that you have a continuous map from $S^2$ to $S^2$ of degree $m$. This is precisely the idea described in both answers to the duplicate question Jason found. – Jonas Kibelbek Mar 01 '12 at 21:15
  • @Jonas, thank you. I find the construction using suspension interesting. – Herband Mar 01 '12 at 21:30
  • @math: What part of the duplicate are you having trouble understanding? Perhaps you can ask a question about exactly where you're having trouble. – Jason DeVito - on hiatus Mar 01 '12 at 22:59
  • @JasonDeVito, I have already understood the suspension construction from Hatcher's book. Thank you! – Herband Mar 02 '12 at 15:20

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