# Odd degree and antipodal points

I am looking at Problem 7 in Milnor's Topology from the Differentiable Viewpoint, namely :

Show that any smooth map $S^n \to S^n$ of odd degree must carry some pair of antipodal points into a pair of antipodal points.

Here is my attempt : consider the contraposition and take $f : S^n \to S^n$, smooth, such that for all $x \in S^n$, $$f(x) \neq -f(-x)$$ (this is saying that no pair of antipodal points is carried to another pair of antipodal points by $f$). We will try to show that $deg(f)$ is even.

Another way to interpret this is to say that for all $x \in S^n$,

$$||f(x)-f(-x)||<2$$ as the equality only holds if $f(x)$ and $f(-x)$ are antipodal. Thus we can build a smooth homotopy $F$ between $f(x)$ and $f(-x)$ (cf Problem 3), given for example by $$F(x,t)=\frac{tf(x)+(1-t)f(-x)}{||tf(x)+(1-t)f(-x)||}$$ Now if $f(x)$ and $f(-x)$ are smoothly homotopic, then they must have the same degree. Rewriting $f(-x)$ as $f(a(x))$, where $a : S^n \to S^n$ is the antipodal map, we have that $$deg(f)=deg(f\circ a)=deg(f)deg(a)=(-1)^{n+1}deg(f)$$ If $n$ is even, then $deg(f)=0$. However is $n$ is odd we cannot conclude with this method. Is there something obvious that I am missing, or is this the wrong approach ? Any small tip would be appreciated.

I have also proven that (cf Problem 6) if $g : S^n \to S^n$ is smooth and such that $deg(g) \neq (-1)^{n+1}$, then $g$ must have a fixed point. In our case we must have $deg(f)\neq(-1)^{n+1}$, so $f$ has a fixed point.

• As @R. Alexandre suggests, you really want to be concentrating on the map $g(x) = \dfrac{f(x)+f(-x)}{\|f(x)+f(-x)\|}$, and you can easily show that $\deg_2 g = 0$. – Ted Shifrin Apr 9 '17 at 16:35
• Yes, I read the answer without even trying to figure out what $F(x,1/2)$ would look like (numbers ? Ew !). It's much clearer now. Thank you for your comment. – Bass Apr 9 '17 at 16:50

The Brouwer theorem (p. 51) says that two functions $g,h:S^n\to S^n$ have same degree iff they are (smoothly) homotopic. So $f(x)$ and $f(-x)$ can't be smoothly homotopic if $n$ is odd.

So you "really" need a new idea to finish your proof for the case $n$ odd.

(Sorry I don't have any tip to give you right now, I need to think a bit. And since I don't have enough reputation I can't just post this as a comment...)

edit :

Ok I think I have a nice clue for you. You should compare degrees of $F(x,t)$ for $t=0$ and $t=1/2$ ;)

• Thank you for your remark. However I don't really see how the clue relates to the problem... I'm not even sure I can say anything about $F(x,1/2)$ ! – Bass Apr 9 '17 at 14:37
• Try to count the number of preimages. They come by pairs. So the degree is even for t=1/2 – R. Alexandre Apr 9 '17 at 14:53
• This is great. Thanks ! – Bass Apr 9 '17 at 16:47