Let $M$ be a smooth manifold of dimension $n$ and let $S$ be an embedded submanifold of $M$, i.e. a subset of $M$ which is given a structure of smooth manifold of dimension $k\leq n$ such that the inclusion map $\iota:S \to M$ is a smooth embedding.
By the constant rank theorem, if $p$ is a point of $S$, then there exists smooth charts $(W,\chi)$ for $S$ centered at $p$ and $(V,\psi)$ for $M$ centered at $p$ with $W \subseteq V$ such that the map $\psi\circ\chi^{-1}:\chi(W)\to\psi (V)$ has the expression $(x^1,\dots,x^k)\mapsto (x^1,\dots,x^k,0,\dots,0) $.
In particular the map $\psi\circ\chi^{-1}:\chi(W)\to\psi (W)$ is a bijection and thus $$\psi(W)=\chi(W)\times\{ \underline{0} \}$$
Since $W$ is open in $S$, we have $W=S\cap A$ with $A$ open subset of $M$.
Let be $U=V \cap A$. Then $U$ is open in $V$ and thus $\psi(U)$ is open in $\mathbb{R}^n$. We also have $U\cap S=W$. So we have $$\psi(U\cap S)=\chi(W)\times\{ \underline{0} \}\quad [1]$$
I want to say that $\psi(U\cap S)=\{ x\in\psi(U) : x^{k+1}=\dots=x^n=0\}$.
My notes say that I can suppose $\psi(U)=B^n_\varepsilon(0)$ and $\chi(W)=B^k_\varepsilon(0)$ be open balls in $\mathbb{R}^n$ and $\mathbb{R}^k$ respectively with the same radius but I can't see why.
I know that I can WLOG assume $\chi(W)=B^k_\varepsilon(0)$ for a certain $\varepsilon>0$ but then $U$ is determinated by $W$! (See the definition of $U$) So I can't modify $U$ and then still pretend that $[1]$ holds.
Can you tell me why the conclusion of my notes is true (if it is true!) or provide another strategy to conclude the argument?
Best Answer
Let us understand what the claim is doing here.
So what we want to show is thatany embedded manifold satisfies the condition.
Your set equality is by definition: but let me explain the steps.
Making the radius smaller is used in first changing your $W$ and $V$. Choose open balls $B' \subseteq V$, $B \subseteq W$, such that the image of $B$ is precisely a $k$-sliceof $B'$. This is possible by making the radius of the balls sufficiently small. We may take these as our $W$ and $V$.
One might think we are done here. That $V$ is our desired $k$-slice. i.e. $V \cap S=W$. Nonetheless, we have: $$ W \subseteq V \cap S$$
The remedy is to introduce your open set $A$.
As you wrote, $W = S \cap A$ for an open set $A$ in $M$. Then $U= A \cap V$ is our desired $k$-slice. This is a set chase. One direction is already proven. Now $$ U \cap S = (A \cap V) \cap S = (A \cap S) \cap V = V \cap W = W$$
In words, this means,
At this point it is clear, in coordinates: if $(x,0) \in U \cap S$, then it lies both in $S$ and is in the $k$-slice of $U$. But above shows this implies it must lie in $W$. Hence $U \cap S \subseteq W$. The converse is also clear, we know $W$ is represented as the $k$-slice in $U \cap S$. Thus your equality.
In fact more is true. If a subset satisfies the condition, then it can be given a (unique) manifold structure such that it is an embedded $k$-manifold.