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Orthogonal projection $S^{2} \to R^{2}$ has exactly one antipodal pair with equal image

Example

For the coordinate projection

$q : S^{2} ⟶ R^{2}, q (x, y, z) = (x, y),$

the only antipodal pair with equal image is the unordered north-south pair ${(0, 0, 1), (0, 0, - 1)}$ .

Facts & Assumptions

Given: The unit sphere $S^{2} \subseteq R^{3}$ and the coordinate projection $q (x, y, z) = (x, y)$ .

[L1]

Every continuous map $S^{2} \to R^{2}$ has an antipodal pair with equal image (Borsuk–Ulam theorem in dimension two).

[F1]

The sphere $S^{2}$ consists of triples $(x, y, z)$ satisfying $x^{2} + y^{2} + z^{2} = 1$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[L2]

Continuity of a map into $R^{2}$ is equivalent to continuity of its coordinate functions (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).

Verification

technique · direct

1.1

Both coordinate functions of $q$ are continuous, so $q$ is continuous by [L2].

givenL2

2.1

If $v = (x, y, z) \in S^{2}$ satisfies $q (v) = q (- v)$ , then $(x, y) = (- x, - y)$ and hence $x = y = 0$ . The unit-sphere equation gives $z^{2} = 1$ , so $v = (0, 0, 1)$ or $v = (0, 0, - 1)$ . These are two points forming exactly one antipodal pair, and they do have equal image $(0, 0)$ , in agreement with [L1].

step 1.1F1L1algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources

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

Orthogonal projection S2→R2S2→R2 has exactly one antipodal pair with equal image