# Is an immersion an embedding almost everywhere?

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?

• To those who voted to close: could you kindly share some input on how to improve the question?
– user242708
Feb 19, 2022 at 18:27
• 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. Mar 14, 2022 at 0:52

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.
• @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). Feb 20, 2022 at 12:24