I am trying to solve the problem A map without fixed points - two wrong approaches. But I am not certain about the degree of antipodal map.

I my thought, since the preimage of a point $y \in S^k$ is just $-y$, the degree is just $+1$ or $−1$, depending on the orientation of $-y$?

  • $\begingroup$ What is "orientation of -y" supposed to mean?? Be rigorous please. The hint is to write down the explicit antipodal map in coordinates and view it in terms of reflections. Then you're reduced down to "what is the degree of a reflection map". $\endgroup$ – Chris Gerig Jul 26 '13 at 18:42


The degree of a reflection through a hyperplane is $-1$, because it is an orientation reversing diffeomorphism. The antipodal map on $S^k$ can be written as the composition of $k+1$ reflections through hyperplanes.

Now use the following very nice fact about the notion of degree of a continuous map. Given any continuous $f,g:S^k\to S^k$, the degree of the composition of $f$ and $g$ is the product of the degrees. In other words:

$$\deg(f\circ g)=\deg f\cdot\deg g$$

  • $\begingroup$ Hi Jared, in fact I think I got it! Thank you so much for your clear guidance! However, I am not able to get through the fundamental fact that the degree of a reflection through a hyperplane is −1, though it appears certainly true. In my book, it is defined to be the sum of preimage orientation, but I was not able to work it out! Can you give me a bit guidance? Thank you so much! math.stackexchange.com/questions/453056/preimage-orientation $\endgroup$ – WishingFish Jul 26 '13 at 23:27
  • $\begingroup$ Thanks for regurgitating my hint... $\endgroup$ – Chris Gerig Jul 27 '13 at 20:58
  • $\begingroup$ Still can't carry out the proof of degree of a reflection, Jared...math.stackexchange.com/questions/453875/… @ChrisGerig $\endgroup$ – WishingFish Jul 28 '13 at 5:18

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