CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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There is no continuous injection from $S^{2}$ into $R^{2}$

Statement

There is no continuous injective map $S^{2} \to R^{2}$ .

Facts & Assumptions

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

[L1]

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

[F1]

The unit sphere $S^{2}$ consists of the vectors $x \in R^{3}$ with $∥ x ∥_{2} = 1$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[F2]

A map is injective when equality of two images forces equality of their inputs (Injection, surjection, bijection).

Proof

technique · direct

1.1givenL1choose

By [L1], choose $x \in S^{2}$ with $f (x) = f (- x)$ .

2.1step 1.1F1F2algebra∎

If $x = - x$ , then $2 x = 0$ and hence $x = 0$ , contrary to $∥ x ∥_{2} = 1$ . Thus $x$ and $- x$ are distinct points with the same image, so $f$ is not injective.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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, consequence after Theorem 1.10 (standard reference, not scraped)

There is no continuous injection from S2S2 into R2R2