Smooth functions $\mathbb{S}^1 \rightarrow \mathbb{R}^2$ I'm reading some lecture notes on differential topology, and it is said that any smooth function $\mathbb{S}^1 \rightarrow \mathbb{R}^2$ whose image is the figure "8" is an immersion but not an embedding.
I tried to prove it on my own, but I couldn't find any good idea; could you give me a hint?
Also, for the "embedding part", I suppose the idea is to prove that the fiber of the double point is not a singleton. Is the continuity sufficient to prove this, or the smoothness is strictly necessary?
 A: First let's talk about what's happening concretely. Here's a figure $8$ that we can draw in desmos:

This is given by the function
$$
\begin{align}
f : S^1 &\to \mathbb{R}^2 \\
\theta &\mapsto \big ( \!\sin(2\theta), \cos(\theta) \big )
\end{align}
$$
Now, it's easy to see that this is not an embedding. After all, $f ( \pi / 2) = f (3 \pi / 2) = (0,0)$, so $f$ is not injective.
But even though it's not an embedding, it's still super easy to see that the double point at the origin really does come from two points in $S^1$. Why? Because the tangent spaces are different. For $\theta = \pi/2$, the tangent space is "sloping up"

while for $\theta = 3 \pi / 2$, the tangent space is "sloping down"

I'll leave it to you to do the quick computation to verify these (as a hint, the tangent space at $\pi/2$ is the span of $(-2,-1)$, while the tangent space at $3 \pi / 2$ is the span of $(-2,1)$), but you can see that we recover injectivity on tangent spaces.
After all, the tangent space of every point in $S^1$ is $\mathbb{R}$, and for each point we're (injectively) sending that tangent space to a one dimensional subspace of $\mathbb{R}^2$.

Algebra and calculus are great, and it's important to be able to do these kinds of computations. But always remember they're supposed to mean something. So is there some way we could geometrically see that the figure 8 is an immersion but not an embedding? Then we should think of the computation we just did as making our intuition precise.
Unsurprisingly, the answer is "yes", and it's already implicit in the way I was talking about the tangent spaces earlier!
You should really think of the figure 8 not as its graph, but rather as the function $f$. If you're familiar with parametric equations, we can get some intuition by thinking about $f$ as describing a point in space at time $\theta$.
Then it's obvious from the picture that we don't have an embedding. After all, in the process of drawing the figure 8 (which you should think of as following the particle around the figure 8) we have to cross over a point we've already drawn. That is, our function cannot be injective.
It's also obvious from the picture that we do have an immersion, since even though we crossed ourselves, the direction we're traveling is always well defined. That is, we always have a nonzero tangent vector, so our tangent space (inside $\mathbb{R}^2$) is always one dimensional (the same dimension as the tangent space in $S^1$).

As a last aside, there are, of course, higher dimensional analogues of this phenomenon. For instance, the klein bottle immerses in $\mathbb{R}^3$

This is not an embedding, since we have self intersection. But it is an immersion. You should convince yourself of this by thinking about the tangent space of the klein bottle (this will be $\mathbb{R}^2$ at each point), and noticing that at each point of the image we also have a well defined full 2-dimensional tangent plane.
Just like in the figure 8 example, we have two such tangent planes at the point of intersection (compare with having two tangent lines at the point of intersection of the figure 8), but since these come from two different points inside the klein bottle (compare with two different points inside $S^1$), and this is ok.

I hope this helps ^_^
