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I have a hard time understanding the concept of immersions. In my course, it was only introduced by the immersion theorem wich says:

Let $f: U \subset \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$ be $C^{1}$ on an open subset $U \subset \mathbb{R}^{m}$. Let $a \in U$ rank$(Df_{(a)}) = m$ then there exists a neigbourhood $V$ of $f(a)$ and a diffeomorphism: $\psi: V \rightarrow \psi(V)$ such that: $$\psi ( f(x_{1},...,x_{n})) = (x_{1},...,x_{n},\underbrace{0,...0}_\text{n-m})$$ For $x$ close enough to $a$.

The way I see it is that a 'surface' in $\mathbb{R}^{m}$ can be gradually deformed to a flat surface in $\mathbb{R}^{n}$. Is this a reasonable way of thinking of immersions? If so, consider: $$f:\bar{B}_{1}(0) \subset \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}:(x,y) \rightarrow (x,y,\sqrt{1-x^{2}-y^{2}})$$ wich maps $\{ (x,y) \in \mathbb{R}^{2}| x^{2}+-y^{2} \leq 1 \}$ to the upper halve of a sphere in $\mathbb{R}^{3}$. The differential of $f$ in $(0,0)$ is: $$Df_{0} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}$$ So the immersiontheorem says that in a neighbourhood of the 'north pole' of the upper sphere the surface can be flattened by a function $\psi$. Consider now the function:

$$g:\mathbb{R}^{2} \rightarrow \mathbb{R}^{3}: (\theta,\phi) \rightarrow (\sin\theta \cos \phi,\sin\theta \sin \phi, \cos \theta)$$ Wich also represents a sphere. (not injective). The differential in $(0,\phi)$ is now: $$Dg_{0} = \begin{pmatrix} \cos \phi & 0\\ \sin \phi & 0 \\ 0 & 0 \end{pmatrix}$$

In this case rank$\left (Dg_{(0,\phi)} \right)<2$ so the immersion theorem doesn't apply. My feeling is that $g$ can't be flattened near $(0,\phi)$ because the immersion theorem doesn't say it can be done. Is this true (I'm aware of the immersion theorem not being if and only if)? and why is it? Because the surface of $g$ and $f$ are the same near those points.

Also: What does immersions make important?

Bear in mind that I haven't learned anything of manifolds. I looked at another question , but I don't have enough knowledge to understand it.

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    $\begingroup$ An immersion is locally (in the domain) an embedding, but not necessarily globally. If you can't get an actual embedding $M \hookrightarrow N$, an immersion is the next best thing. $\endgroup$ – user98602 Mar 31 '15 at 19:33
  • $\begingroup$ So the idea is to have a function that does locally have the properties of a diffeomorphism (only not surjective) ? $\endgroup$ – abcdef Mar 31 '15 at 19:41
  • $\begingroup$ As in the theorem you have mentioned, an immersion is like a projection. $\endgroup$ – Eclipse Sun Apr 5 '15 at 15:06
  • $\begingroup$ That's sounds more like submersion. Projection is like $(x,y) \rightarrow x$. Do you mean injections like: $x \rightarrow (x,0)$? $\endgroup$ – abcdef Apr 5 '15 at 21:36
  • $\begingroup$ Sorry, I misunderstood the question. I should say immersion is 'dual' to submersion. It behaves locally like embedding. $\endgroup$ – Eclipse Sun Apr 6 '15 at 4:51
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The main point with inmersions is how they relate to embeddings. The latter are diffeomorphic copies: $h:M\hookrightarrow\mathbb R^n$ is an inmersion if it gives a diffeomorphism from $M$ onto $N=h(M)\subset\mathbb R^n$ and $N$ is a submanifold of $\mathbb R^n$ (one could use any other manifold instead of $\mathbb R^n$). Thus there is no diff-top distinction among the initial $M$ and the copy $N=h(M)$. Now for inmersions this is true only locally at the source: If $f:M\to\mathbb R^n$ is an inmersion at $a\in M$ say, there is an open nbhd $U$ of $a$ in $M$ such that $f|_U:U\to\mathbb R^n$ is an embedding, that is, $U$ is diffeomorphic to $f(U)$, which is a submanifold of $\mathbb R^n$. And here comes the important thing to notice: $f(U)$ need not be open in $f(M)$, hence we cannot deduce $f(M)$ is a manifold near $f(a)$. This obstruction occurs even if $f$ is injective. This is the famous example:

lemniscate

(which I borrow from Find $f:C\to\mathbb{R}^2$ continuous and bijective but not open, $C\subset\mathbb{R}^2$ is closed). Here $$ f:\mathbb R\to\mathbb R^2:t\mapsto\big(\frac{t^3}{1+t^4},\frac{t}{1+t^4}\big). $$ The image $f(\mathbb R)$ is a lemniscate with two branches crossing at the origin $(0,0)=f(0,0)$, at which $f(\mathbb R)$ is not a regular curve. But if you consider a small open interval $U$ of $0\in\mathbb R$, the image is a small open piece of the vertical branch which is indeed a regular curve, but is not open in the lemniscate (it misses the horizontal branch).

This example stress the fact that an imersion is an embedding iff it is a homeo onto its image. This is important to understand and visualize projective plane and in general compact non-orientable surfaces. These surfaces can be embedded into $\mathbb R^4$ but not in $\mathbb R^3$. Since the next best thing to an embedding is an immersion, one looks at immersions into $\mathbb R^3$. For the projective plane there are several classical ones with famous names: the Steiner embeddings (like the roman surface or the crosscap), or the Boy surface. They have self-intersections and singular points of course. A search in the web gives nice pictures of them all. as well as some of the standard immersion of the Klein Bottle in $\mathbb R^3$ (which justifies its name), which can even be seen in some beautiful glass models.

Another important class of immersions that are not embeddings are some "parametrizations" of manifolds, as the one of the sphere in the question. These are immersions whose image is in fact a smooth manifold. The are not homeos onto its image, but topological identifications. They give local parametrizations of that manifold when restricted to suitable open sets. Typical examples are the classical parametrizations of spheres, or tori.

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  • $\begingroup$ Thanks, do you know why the immersion doesn't applies for the example function $g$? $\endgroup$ – abcdef Apr 12 '15 at 12:20
  • $\begingroup$ This example $g$ is a parametrizatin of the sphere, hence the image is in fact a submanifold. Here the only obstruction to being an embedding is injectivity. The best one can get is an embedding when restricted to an open set whose image is the sphere except some half-meridian. You can restrict to a subset on which injectivity holds and whose image is the whole sphere, but the subset is not open. So, $g$ is an immersion, not an embedding. $\endgroup$ – Jesus RS Apr 12 '15 at 17:27

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