Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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FALSE: every injective immersion is a proper embedding

Statement

False claim: every injective immersion is a proper embedding.

Facts & Assumptions

Given: The map f:(0,1)R2,f(t):=(cos2πt,sin2πt).

[L1]

A proper injective immersion is a smooth embedding (A proper injective immersion is a smooth embedding).

Refutation

technique · direct
1.1

The map f is smooth and injective, and its derivative f(t)=(2πsin2πt,2πcos2πt) never vanishes. Thus f is an injective immersion.

givenalgebra
2.1

The compact set K:=S1R2 has inverse image (0,1), which is not compact. So f is not proper. Its image is also not closed in R2.

step 1.1algebra
3.1

By [L1], properness is exactly the extra hypothesis missing from this example. Therefore the displayed injective immersion is not a proper embedding, and the claim is false.

L1step 2.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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