Let $M, N$ be smooth manifolds, and let $F \colon M \to N$ 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?

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    $\begingroup$ To those who voted to close: could you kindly share some input on how to improve the question? $\endgroup$
    – user242708
    Feb 19, 2022 at 18:27
  • $\begingroup$ Can only guess, but it is really nothing more than a problem statement question (PSQ), no better than someone writing the preamble of a question in a Calc I text, then asking what the exercise asks of them. $\endgroup$
    – amWhy
    Mar 14, 2022 at 0:52

2 Answers 2


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 $S^1=\{z\in\mathbb{C}\mid |z|=1\}$ to itself by $z\mapsto z^2$. 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 ($A(z)=-z$) then $T\cap A(T) = \emptyset$, so that $\lambda(T) + \lambda(A(T)) \le \lambda(S^1)$, where $\lambda$ is the length measure on $S^1$; but $\lambda(T)=\lambda(A(T))$, since $A$ is an isometry.


In case it's of interest, even an injective immersion can fail everywhere to be an embedding. An irrational winding on a torus is an example: Fix an irrational number $\alpha$. The path $\gamma(t) = (t, \alpha t)$ in the real plane descends to an injective regular path on the square torus $(\mathbf{R}/\mathbf{Z})^{2}$ whose image is dense.

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    $\begingroup$ I don't see how density implies nonzero measure. The set of rational numbers is dense in the reals, but has measure zero. $\endgroup$ Feb 20, 2022 at 6:14
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    $\begingroup$ @Acccumulation You're right that density does not imply positive measure, but that's not the issue here: The image of the reals ($M$) is dense in the torus ($N$), so the irrational winding fails to be an embedding at every real number (i.e., $X = M$ is the entire domain). $\endgroup$ Feb 20, 2022 at 12:24

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