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Borsuk–Ulam theorem in dimension two

Statement

For every continuous map $f : S^{2} \to R^{2}$ , there is an $x \in S^{2}$ with $f (x) = f (- x)$ .

Facts & Assumptions

Given: A continuous map $f : S^{2} \to R^{2}$ .

[F1]

The sphere $S^{2}$ is the unit sphere in $R^{3}$ , and its equator is the image of $e : R / Z \to S^{2}$ , $e ([t]) = (\cos 2 π t, \sin 2 π t, 0)$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ , $[t] \mapsto (\cos 2 π t, \sin 2 π t)$ is a homeomorphism from $R / Z$ to the unit circle).

[L1]

Every continuous antipodal map $S^{1} \to S^{1}$ has an odd lift increment and is not nullhomotopic (An antipodal circle map has odd lift increment and is not nullhomotopic).

[L2]

The sphere $S^{2}$ is simply connected ( $S^{n}$ is simply connected for every $n \geq 2$ ).

[L3]

Radial normalization $ρ : R^{2} ∖ {0} \to S^{1}$ , $ρ (y) = y / ∥ y ∥_{2}$ , is continuous (Radial normalisation $x \mapsto x / ∥ x ∥_{2}$ is continuous on $R^{n} ∖ {0}$ ).

[L4]

Postcomposition by a continuous map preserves a homotopy relative to its fixed subspace (Precomposition and postcomposition by continuous maps preserve homotopies, including their relative form).

[L5]

Continuity of maps into Euclidean space is componentwise, and sums and scalar multiples of continuous Euclidean-valued maps are continuous (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

Proof

technique · contradiction

1.1

Suppose $f (x) \neq f (- x)$ for every $x \in S^{2}$ .

givenassume-contra

2.1

The difference $d (x) = f (x) - f (- x)$ is continuous by [L5] and nonzero by step 1.1, so $g (x) = ρ (d (x))$ defines a continuous map $g : S^{2} \to S^{1}$ . Since $d (- x) = - d (x)$ , one has $g (- x) = - g (x)$ .

step 1.1L3L5constructalgebra

3.1

Let $h : R / Z \to S^{1}$ be the homeomorphism in [F1] and put $b = h^{- 1} \circ g \circ e$ . The map $e$ is continuous componentwise by [L5]; since $e ([t + 1 / 2]) = - e ([t])$ and $h ([u + 1 / 2]) = - h ([u])$ , the continuous map $b$ is antipodal. Hence the loop $t \mapsto b ([t])$ is not nullhomotopic by [L1].

step 2.1F1L1L5

3.2

The loop $t \mapsto e ([t])$ in $S^{2}$ is nullhomotopic because $S^{2}$ is simply connected. Postcomposing such a nullhomotopy with the continuous map $h^{- 1} \circ g$ makes $t \mapsto b ([t])$ nullhomotopic in $R / Z$ .

step 2.1F1L2L4

4.1

Steps 3.1 and 3.2 contradict one another. Therefore the assumption in step 1.1 is false, and some $x \in S^{2}$ satisfies $f (x) = f (- x)$ .

step 1.1step 3.1step 3.2discharge-contradiction∎

Depends on

Used by

Dependency tree · two levels

57 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Allen Hatcher, Algebraic Topology, Theorem 1.10 (standard reference, not scraped)