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Sonia Gay

Sonia Gay

Answered question

2022-07-01

Let M , N be smooth manifolds, and let F : M N be a smooth immersion. I know that F is a local embedding, but is it also an embedding almost everywhere?
In other words, does there exists a set of measure zero X M such that F restricted to M X is a smooth embedding?

Answer & Explanation

Odin Jacobson

Odin Jacobson

Beginner2022-07-02Added 17 answers

No: a covering map of smooth manifolds is an immersion but (I'm nearly sure) never has this property unless it is a diffeomorphism.
For example, map S 1 = { z C | z | = 1 } to itself by z z 2 . This is an immersion but a measurable subset T of the domain on which the map is injective has measure at most 1 2 the measure of S 1 , because if A : S 1 S 1 is the antipode ( A ( z ) = z) then T A ( T ) = , so that λ ( T ) + λ ( A ( T ) ) λ ( S 1 ), where λ is the length measure on S 1 ; but λ ( T ) = λ ( A ( T ) ), since A is an isometry.

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