A map $f:S^n\to S^n$ of odd degree must carry some pair of antipodal points into a pair of antipodal points

Question

I want to prove by contradiction and hence assume that $f$ does NOT carry any pair of antipodal points into a pair of antipodal points. A smooth homotopy $F(x,t):S^n\times[0,1]\to S^n$ is well defined by $$F(x,t)=\frac{tf(x)+(1-t)f(-x)}{ ||tf(x)+(1-t)f(-x)|| }$$ since each segment connecting $f(x)$ with $f(-x)$ does not contain the origin. By the homotopy invariance, $f(-x)$ has the same degree as $f(x)$, that is, $\deg(f(-x))$ is odd. I was told to take a look at the map $g:S^n\to S^n$ defined by $g(x)=\frac{1}{2}[f(x)+f(-x)]$. OK, it's an even function, and a further study gives that its degree must be even. But I'm still unable to reach a contradiction. Does anyone have an idea? Can we say anything about a linear combination of maps with odd degree? Thanks.

---

Answer 1

The question was asked with differential topology tag so one standard definition for degree(in differential topology) is: If $f$ is a smooth map whose domain is a compact manifold and p is a regular value of $f$, consider the finite set

$${\displaystyle f^{-1}(p)=\{x_{1},x_{2},\ldots ,x_{n}\}\,.}$$ By $p$ being a regular value, in a neighborhood of each $x_i$ the map $f$ is a local diffeomorphism (it is a covering map). Diffeomorphisms can be either orientation preserving or orientation reversing. Let $r$ be the number of points $x_i$ at which $f$ is orientation preserving and $s$ be the number at which $f$ is orientation reversing. When the domain of $f$ is connected, the number $r − s$ is independent of the choice of $p$ (though $n$ is not!) and one defines the degree of $f$ to be $r − s$.

After building some common ground lets proceed with the proof.

Now looking at $g(\cdot)=F(\cdot,\frac{1}{2}):S^n \to S^n$ you defined we observe that $g(x)=g(-x)$. We now have: $$deg(g)=deg(g)\cdot \deg(-I)$$
We observe that in even dimension $-I$ is orientation reversing so $deg(g)=0$ a contradiction to the fact that its homotopic to the odd degree map $f$.

In odd dimension the real projective space $\mathbb{R}P^n$ is orientable. Now we can use the fact that $g$ factors through $\mathbb{R}P^n$. $$S^n \to \mathbb{R}P^n \to S^n$$ where the first arrow is the covering map $\pi:S^n \to \mathbb{R}P^n$ of the projective space. Pick $[p] \in \mathbb{R}P^n$ then $\pi^{-1}([p])=\{p,-p\}$ and as its a covering map locally the map is the identity diffeomorphism hence the degree of $\pi$ is $2$ but this shows that $$deg(g)=deg(\pi)\cdot \deg(\tilde{g})=2\deg(\tilde{g})$$ This shows that $deg(g)$ is even. As before this is a contradiction to the fact that $g$ is homotopic to $f$