We start from the fact that any maps $f,g:S^n\to S^n$ such that $f(x)\neq g(x)$ for any $x$, satisfy $f\simeq a\circ g$ with $a$ the antipodal $a(s)=-s$, that homotopical maps have the same degree, and that $\deg (g\circ f)=\deg g\deg f$ for any $f,g:S^{n}\to S^n$. We also know that $\deg a=(-1)^{n+1}$. From this we can prove the following (From Greenberg and Harper's book on algebraic topology).
Exercise 16.8. Let $f:S^n\to S^n$ be homotopic to a constant. Then $f$ has a fixed point and also a point $x$ at which $f(x)=-x$.
Proof. Since $f$ is homotopic to a constant, then $\deg f=0$, so $f$ must have a fixed point. Now take $g=a\circ f\simeq a\circ \text{cst}_{s_0}=\text{cst}_{-s_0}$ for some $s_0\in S^n$. By the same argument, $g$ has a fixed point, i.e. some $x$ such that $g(x)=-f(x)=x$, but this means $f(x)=-x$.
Exercise 16.9. Any map $f:S^{2n}\to S^{2n}$ has a fixed point or sends some point into its antipode.
Proof. Suppose $f$ doesn't have either a fixed point and a point sent into its antipode. Then $f\simeq a$, and since $\deg a=-1$, $\deg f=-1$. Also, since there are not points $x$ such that $f(x)=-x=a(x)$, $f\simeq a\circ a=1_{S^{2n}}$ so $\deg f=1$. This is a contradiction.
Exercise 16.12. Every map $\bR P^{2n}\to \bR P ^{2n}$ has a fixed point.
Proof. Take $f:\bR P^{2n}\to \bR P^{2n}$. Then $f$ lifts to $\tilde{f}:\bR P^{2n}\to S^{2n}$. Take $\overline{f}=\tilde f\circ \pi:S^{2n}\to S^{2n}$ where $\pi:S^{2n}\to \bR P^{2n}$ is the canonical projection. Then, by 16.9, $\overline{f}$ has a fixed point or a point $x$ such that $f(x)=-x$. In the first case, $\tilde{f}([x])=x$ and in the second one $\tilde{f}([x])=-x$ where $[x]={x,-x}\in \bR P^n$. But this means that $f([x])=\pi\circ \tilde{f}([x])=[x]$, so $[x]$ is a fixed point.
Exercise 16.13. Prove that the projection $\pi:S^n\to \bR P^n$ is not nullhomotopic.
Proof. Let's suppose it's nullhomotopic and $\sigma:S^n\to S^n$ a lift of $\pi$, which is, by . Then $\sigma$ is nullhomotopic by the lifting criterion (If $H$ is a homotopy between $\pi$ and the constant map and since $\pi$ lifts to $\sigma$, then $H$ lifts to a homotopy between $\sigma$ and a lift to the constant map, which is itself constant). Then, by Lemma 1234567, and the fact that $S^{2n}$ is simply connected, $\sigma$ is a deck transformation, which, being surjective, cannot be nullhomotopic, a contradiction.
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