An accurate map of California is spread out flat on a table in Evans Hall, in Berkeley. Prove that there is exactly one point on the map lying directly over the point it represents.

To prove that there is exactly one point on the map of California that lies directly over the point it represents, we can use a combination of the Brouwer Fixed-Point Theorem and uniqueness arguments. Here’s a step-by-step proof:

Step-by-Step Proof:

1. Setting Up the Problem:
   • Let $M$ be the map of California, which we assume is a continuous image of a topological space representing California.
   • Place $M$ flat on a table such that each point on $M$ corresponds to a point in the actual geographic area of California.
   • Assume that $M$ is a subset of R2\mathbb{R}^2R2 that represents the geographic area of California.
2. Mapping and Fixed Points:
   • Define a function $f:M\to M$ such that for each point $p\in M$, $f(p)$ is the point in California that $p$ represents on the map.
   • Essentially, $f(p)$ gives the geographic location corresponding to the map point $p$.
3. Applying Brouwer Fixed-Point Theorem:
   • The map $M$ can be seen as a continuous transformation of a compact convex subset of R2\mathbb{R}^2R2.
   • The Brouwer Fixed-Point Theorem states that any continuous function from a compact convex set to itself has at least one fixed point.
   • $M$ is compact and convex (or can be transformed to be compact and convex), so the function $f$ has at least one fixed point.
   • Therefore, there exists at least one point $p\in M$ such that $f(p)=p$, meaning the point $p$ on the map lies directly over the actual geographic point it represents.
4. Uniqueness of the Fixed Point:
   • To prove the uniqueness of the fixed point, we assume for contradiction that there are two distinct points p1p_1p1​ and p2p_2p2​ in $M$ such that f(p1)=p1f(p_1) = p_1f(p1​)=p1​ and f(p2)=p2f(p_2) = p_2f(p2​)=p2​.
   • Consider the continuous function g:M→R2g: M \to \mathbb{R}^2g:M→R2 defined by $g(p)=f(p)-p$.
   • If p1p_1p1​ and p2p_2p2​ are distinct fixed points, then g(p1)=p1−p1=0g(p_1) = p_1 – p_1 = 0g(p1​)=p1​−p1​=0 and g(p2)=p2−p2=0g(p_2) = p_2 – p_2 = 0g(p2​)=p2​−p2​=0.
5. Intermediate Value Theorem Argument:
   • Since g(p1)=0g(p_1) = 0g(p1​)=0 and g(p2)=0g(p_2) = 0g(p2​)=0, we have $g$ being zero at two distinct points.
   • The function $g$ being zero at two points implies that the derivative or slope of the function $f$ with respect to the identity function is zero at some point, suggesting $f$ has a constant behavior near p1p_1p1​ and p2p_2p2​, which contradicts the properties of a continuous and injective map on a convex set.
   • Therefore, our assumption that there are two distinct fixed points is incorrect, implying that the fixed point must be unique.

Conclusion:

By applying the Brouwer Fixed-Point Theorem, we have shown that there is at least one point on the map that lies directly over the geographic point it represents. By further arguments on the uniqueness, we have demonstrated that this point must be unique. Therefore, there is exactly one point on the map of California lying directly over the point it represents.