What's the geometrical meaning of immersion?

Question

Does that just mean to different tangent vectors, their images are different tangent vectors?

Answer 1

It is often very useful to have some key examples in mind in order to get a good sense for geometrical meanings of such concepts.

Assuming that the geometrical meaning of embedding is clear (immersion and embedding share the constraint that the dimension of the domain is at most the dimension of the codomain), let's look at some key examples (recalled from John Lee's Introduction to Smooth Manifolds) that distinguish an immersion from an embedding (remember that an immersion is locally an embedding, so the only differences are in some sense 'global' ones):

1) The figure 8 loop in $\mathbb{R}^2$ (immersed via the map $t \mapsto (\cos(\pi/2 + t),\sin(2t))$ for $0 < t < 2\pi$). This immersion of $S^1$ in $\mathbb{R}^2$ fails to be an embedding at the crossing point in the middle of the figure 8 (though, you can check that the map is in fact injective). This causes $S^1$ to not be homeomorphic to its image in the induced topology. However the differential of the map is everywhere injective. Note that the same image could be obtained by the immersion $t \mapsto (\cos(t),\sin(2t))$ however this immersion is not injective, whereas the former is.

2) The immersion of $\mathbb{R}$ in the 2-torus via (for example) the map $q\circ i$ where $i : t \mapsto (t,\pi t)$ sends $\mathbb{R}$ first into $\mathbb{R}^2$, and $q : \mathbb{R}^2 \rightarrow \mathbb{R}^2/\mathbb{Z}^2 \cong \mathbb{T}^2$ is the quotient map to the 2-torus. This map is actually an injection whose image is dense in $\mathbb{T}^2$. It takes a little work to see that the induced topology on the image is not the same as the standard one on $\mathbb{R}$ and so the domain of $q\circ i$ is not homeomorphic to its image under the induced topology so, although the map is injective and has everywhere injective differential, it is only an immersion, not an embedding.

3) Finally, let's consider a map that is not even an immersion. The map sending $\mathbb{R}$ into $\mathbb{R}^2$ via the rule $t \mapsto (t^2,t^3)$ is injective, and its differential is injective everywhere except at $t = 0$. The image of $t = 0$ (where the differential is the zero-map) is a cusp on an otherwise smooth plane curve. So the image here is neither immersed, nor embedded.

These examples illustrate fairly well in low dimension how immersions can fail to be embeddings in general. They illustrate the different ways that the image of a map in the induced topology can fail to be homeomorphic to the domain, even if the map is injective. The types of examples of non-immersions though (like (3)) can, in higher dimensions, exhibit quite a bit more complicated behavior at the points where they fail to be immersed.