If $f\in \mathcal{C}(\mathbb{S}^n,\mathbb{S}^n)$ and $\mathrm{deg}_2(f)=0$, then $f(x_0)=x_0$ and $f(y_0)=-y_0$ for some $x_0, y_0$. 
If $f\in \mathcal{C}(\mathbb{S}^n,\mathbb{S}^n)$ and $\mathrm{deg}_2(f)=0$, where $\mathbb{S}^n=\left \{x\in\mathbb{R}^{n+1}:\|x\|=1\right \}$. Prove that there exist two points $x_0, y_0\in \mathbb{S}^n$ such that $f(x_0)=x_0$ and $f(y_0)=-y_0$.

This is an exercise on the chapter about mod 2 degrees and Borsuk-Ulam theorem in a Differential Topology book (Chinese version). I think that a question I asked before is a special case of this exercise. Because, if $\mathrm{deg}_2(f)=1$, then $f$ must be surjective.
Let $g(x)=f(-x)$. Using smooth approximation of $f$ and the definition of $\mathrm{deg}_2$ for smooth functions through regular values, I think I can prove that $\mathrm{deg}_2(f)=\mathrm{deg}_2(g)$. So, it suffices to find a fixed-point for $f$. But I don’t know how to proceed.
Any help would be appreciated!
 A: This is a community-wiki answer illustrating @Rishi's comments under the question I asked. This is the first time that I write a community-wiki answer, please feel free to edit it if anything is inappropriate.
We prove the exercise by contradiction. Assume that $f(x)\neq x$ for all $x\in\mathbb S^n$, then we prove that $f$ is homotopic to the antipodal map $j:\mathbb S^n\to\mathbb S^n, x\mapsto -x$. Indeed, define
$$G: [0,1]\times \mathbb S^n\to\mathbb R^{n+1},\qquad (t,x)\mapsto tf(x)+(1-t)(-x),$$
then $G(t,x)\neq 0$ for all $(t,x)\in[0,1]\times \mathbb S^n$: if $G(t,x)=0$ then $tf(x)=(1-t)x$, so $t=\|tf(x)\|=\|(1-t)x\|=1-t$, which implies that $t=1-t$ and $f(x)=x$; recall that we have assumed that $f(x)\neq x$ for all $x\in\mathbb S^n$ (Here $\|\cdot\|$ is the Euclidean metric in $\mathbb R^{n+1}$.) Hence, the map
$$F(t,x)=\frac{G(t,x)}{\|G(t,x)\|},\qquad (t,x)\in[0,1]\times \mathbb S^n$$
is a homotopy connecting $f$ and $j$. Since $\mathrm{deg}_2$ is invariant under homotopies, we then have $\mathrm{deg}_2(f)=\mathrm{deg}_2(j)=1$, contradicting with "$\mathrm{deg}_2(f)=0$". Therefore, there exists some $x_0\in\mathbb S^n$ such that $f(x_0)=x_0$.
Similarly, if $f(x)\neq -x$ for all $x\in\mathbb S^n$, then we can prove that $f$ is homotopic to the identity map $i:\mathbb S^n\to\mathbb S^n, x\mapsto x$, which is a contradiction, too. So, there exists some $y_0\in\mathbb S^n$ such that $f(y_0)=-y_0$. This completes the proof of this exercise.
Remark. After learning the concept of (oriented) degrees, the same argument proves the following exercise:

Let $f\in \mathcal{C}(\mathbb S^n,\mathbb S^n)$, then

*

*if $\mathrm{deg}(f)\neq 1$, then there exists some $x_0\in\mathbb S^n$ such that $f(x_0)=-x_0$;

*if $\mathrm{deg}(f)\neq (-1)^{n+1}$, then there exists some $x_0\in\mathbb S^n$ such that $f(x_0)=x_0$.


(Note that $S^n\subset \mathbb R^{n+1}$, so $\mathrm{deg}(j)=(-1)^{n+1}$.)
