Theorem: If $f : S^{2n} \rightarrow S^{2n}$ is a continuous map, then there is an $x \in S^{2n}$ with $f(x) = x$ or $f(x) = -x$.

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Proof: It suffices to show that at least one of $f$ and $-f$ must have a fixed point. If $f$ has a fixed point, then clearly it satisifes $f(x) = x$. If $-f$ has a fixed point, then it satisfies $-f(x) = x \Rightarrow f(x) = -x$.

To that end, suppose towards contradiction that $f$ and $-f$ both have no fixed points. Since $f$ has no fixed points, we have $\deg f = (-1)^{2n+1} = -1$. At the same time, since $-f$ has no fixed points, we have $\deg(-f) = -1$ as well, but:

$$\deg(-f) = \deg (-\id \circ f) = \deg(-\id) \deg(f) = (-1)(-1) = 1$$

and we have a contradiction. So at least one of $f$ and $-f$ must have a fixed point.

Corollary: Every map $f : \mathbb{R}P^{2n} \rightarrow \mathbb{R}P^{2n}$ has a fixed point.

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Proof:

$$S^{2n}$$ is a cover of $$\mathbb{R}P^{2n}$$, and by the lifting lemma $$f$$ lifts to a map $$\wt{f} : \mathbb{R}P^{2n} \rightarrow S^{2n}$$. If $$\rho : S^{2n} \rightarrow \mathbb{R}P^{2n}$$ is the canonical cover, then we can lift the whole map $$f$$ to a map $$g = \wt{f} \rho : S^{2n} \rightarrow S^{2n}$$, as in the diagram: <div class="math-figure"><img src="/assets/math_solutions/hatcher/e2-2-2_1.svg" width="270"/></div> Then, there is an $$x$$ with $$g(x) = x$$ or $$g(x) = -x$$. Then, $$f(\rho(x)) = f([x]) = \rho(g(x)) = [x]$$, since $$\rho(x) = \rho(-x) = [x]$$. Thus, $$f$$ has a fixed point $$[x]$$. We end by demonstrating a map from $$\mathbb{R}P^{2n-1} \rightarrow \mathbb{R}P^{2n-1}$$ that does *not* have a fixed point. First consider a linear transformation $$g : \mathbb{R}^{2n} \rightarrow \mathbb{R}^{2n}$$ without eigenvectors in $$\mathbb{R}^{2n}$$. A perfectly good example is the map $$g$$ defined by the $$2n \times 2n$$ matrix:$$

g = \tmat{0}{-I_n}{I_n}{0}

$$

This map has eigenvalues $\pm i$.

We identify $\mathbb{R}P^{2n-1}$ with the set of lines through the origin in $\mathbb{R}^{2n}$ (i.e. one dimensional subspaces), and show that $g$ induces a well-defined map. Indeed, $g$ has full rank so it certainly sends one-dimensional subspaces to other one-dimensional subspaces, so this is a well defined map $\mathbb{R}P^2 \rightarrow \mathbb{R}P^2$. Since it has no real eigenvalues, none of these one-dimensional subspaces are mapped onto themselves, so this map has no fixed points.