Why is the image of the map $f(x) =(sin(x),1- cos(x))$ for $x\in [0, 2\pi)$ and $f =\operatorname{id}$ otherwise not a submanifold of $\mathbb{R}^2$?

Question

Consider the map $f: (-\infty, 2\pi) \to \mathbb{R}^2$ s.t $f(x) = (sin(x), 1- cos(x))$ for $x\in [0, 2\pi)$ and $f(x) = (\operatorname{id}(x), 0)$ for the rest of its domain.

In the book of Chillingworth, Differential Topology, it is stated that the image of this map is not a sub manifold of $\mathbb{R}^2$ even though it has maximal rank for any point on its domain.

However, I cannot understand why it is the case, i.e why the image is not a sub manifold ?

Answer 1

It is not a submanifold because the region near $(0,0)$ is not locally hoeomorphic to $\mathbb R$. Take any neighborhood of $(0,0)$, remove $(0,0)$ from it, and you will get at least three connected components. If it was locally homeomorphic to $\mathbb R$, then you should get only two connected components in some cases.

Answer 2

A circle together with a half-line going to the left.

The point at the bottom of the circle is your problem, where no neighborhood of it is homeomorphic to $(-\epsilon,\epsilon)$.