Injective immersion and embedding

[Korybut]

TL;DR

Why local closedness matters?

Hello!

There is a simple line in the textbook.

If $S$ is a manifold, an injectively immersed submanifold $M$ of $S$ is embedded if and only if $M$ is locally closed in $S$.

Recall the definition. M is locally closed if for each point $x\in M$ there open $U\subset S$ such that $M\cap U$ is closed in $U$.

Embedding to injective immesion is simple. The opposite direction is hard.

Suppose I have $N$ as source manifold and $f:N\rightarrow S$ is the injective immersion. I need to show that "immersed" topology coincides with subspace topology. In other words: for any open $V \subset N$ I need to show that $f(V)=M\cap U_V$ where $U$ is open in S.

Immersion allows to use constant rank theorem and build chart $(A,\phi)$ with $\phi : A \rightarrow \mathbb{R}^k$ for $A\subset N$; and chart $(B,\psi)$ with $\psi: B\rightarrow \mathbb{R}^{n+k}$ for $B\subset S$. In these charts immersion has the form
$\psi \circ f \circ \phi^{-1} :\mathbb{R}^k \rightarrow \mathbb{R}^{n+k}$
$(x_1,...,x_k)\rightarrow (x_1,...,x_k,0,...0)$.

How do I proceed further?

 

[Korybut]

As it happens usually once you post you are full of ideas

I have the following sketch of the proof. Since $f(N)=M$ is locally closed for any point $x\in M$ there is open $U\subset S$ such that $M\cap U$ is closed in $U$.
By assumption $f:N\rightarrow S$ is continuous. Therefore I define open in $N$ set $V$
$V:=f^{-1}(U)\subset N$
Also we can consider restriction of $f$, namely $f|_V$
$f|_V :V\rightarrow f(N)\cap U$
This map is bijective (since $f$ is assumed injective) and continuous in the direction $V\rightarrow f(N)\cap U$ (by assumption).

Tricky part is to show $f|_V {}^{-1}$ is continuous. I believe this is where that M is locally closed is handy.
Since $V$ and $M\cap U$ are manifolds one consider convergent sequences and prove the continuous property of $f|_V {}^{-1}$ from sequential definition.
Suppose I pick a convergent sequence ${f(p_n)}\in f(M)\cap U$. Since set $f(M)\cap U$ is closed (in $U$) it contains all limit points. So
$\lim_{n\rightarrow \infty} f|_V {}^{-1}(f(p_n))=p$
with some $p$ that comes from bijective property of f. I.e. for the limit $\lim_{n\rightarrow \infty} f(p_n)=\tilde{f}$ there should such $p$ that $\tilde{f}=f(p)$.

Do I use this property of local closedness correctly?

 

[Korybut]

The original statement is false, even though written in a textbook. Counter example is well known actually

Consider the map $(0,2\pi)\rightarrow \mathbb{R}^2$ such that $t\rightarrow (\sin 2t, \sin t)$. Image of this map is figure-eight, which is closed in $\mathbb{R}^2$, map itself is injective and immersion however topology of immersion is finer than subspace topology.

Real criterion for immersion to be embedding is the properness.