Sonia Gay

2022-07-01

Let $M,N$ be smooth manifolds, and let $F:M\to N$ 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 $X\subset M$ such that $F$ restricted to $M\setminus X$ is a smooth embedding?

Odin Jacobson

Expert

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 ${S}^{1}=\left\{z\in \mathbb{C}\mid |z|=1\right\}$ 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 $\frac{1}{2}$ the measure of ${S}^{1}$, because if $A:{S}^{1}\to {S}^{1}$ is the antipode ($A\left(z\right)=-z$) then $T\cap A\left(T\right)=\mathrm{\varnothing }$, so that $\lambda \left(T\right)+\lambda \left(A\left(T\right)\right)\le \lambda \left({S}^{1}\right)$, where $\lambda$ is the length measure on ${S}^{1}$; but $\lambda \left(T\right)=\lambda \left(A\left(T\right)\right)$, since $A$ is an isometry.

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