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.
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 $t=1/2$ ;)