False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-01

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A preimage need not be a submanifold without transversality

Statement

False claim: the preimage of every embedded submanifold under a smooth map is again a submanifold.

Facts & Assumptions

Given: The smooth map $F : R^{2} \to R$ , $F (x, y) = x y$ , and the embedded submanifold ${0} \subseteq R$ .

[L1]

The transverse preimage theorem needs transversality to conclude the preimage is a submanifold (The transverse preimage theorem).

Refutation

technique · direct

1.1

The preimage is $F^{- 1} (0) = {(x, y) : x y = 0} = (R \times {0}) \cup ({0} \times R),$ the union of the two coordinate axes.

givenalgebra

2.1

At the origin this set has two distinct tangent directions, so no neighbourhood of $(0, 0)$ is diffeomorphic to an open interval or to a point. Hence it is not a $1$ -dimensional or $0$ -dimensional embedded submanifold there. This is exactly the failure excluded by the hypothesis in [L1].

L1step 1.1

3.1

Therefore a preimage need not be a submanifold when transversality is dropped.

step 2.1∎

Depends on

The transverse preimage theorem

Used by

A tangent intersection whose set-theoretic intersection is not of the expected dimension Counterexample

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

Marco Gualtieri, Topology I: Smooth Manifolds, cumulative notes (standard reference, not scraped)