Equivalent condition for $f:M\to N$ to be an embedding 
Let $M$, $N$ be smooth manifolds and $f:M\to N$ a smooth, injective map. Prove that $f$ is an embedding if and only if:
$\exists\,k$ such that $\forall q\in f(M),\,\exists\,(V,Y)$ at $q$ such that $Y(V\cap f(M))=Y(V)\cap (\mathbb{R}^k\times\{0\}^{n-k})$

[here, embedding means that $f$ is an immersion and a homeomorphism onto its image]
I'm stuck in both directions:
($\Leftarrow$) Since $f:M\to f(M)$ is obviously continuous bijection, I'm left to prove that $f$ is open, i.e., for any open $U\subset M$, we have $f(U)$ open in $f(M)$, which means $f(U)=W\cap f(M)$ for some open $W\subset N$. I still have no idea how to find such $W$ given the above hypothesis.
($\Rightarrow$) Looks harder, but here's what I've thought: if $U\subset M$ is open, then $f(U)\subset f(M)$ is open, which means $f(U)=V\cap f(M)$ for some open $V\subset N$. Without loss of generality, suppose there are charts $(U,X)$ and $(V,Y)$. Defining $k:=\dim M$, we have that $X(U)$ is an open subset of $\mathbb{R}^k$ with $X(U)\simeq U\simeq f(U)=V\cap f(M)\simeq Y(V\cap f(M))$ ($\simeq$ means homeomorphic), so $Y(V\cap f(M))$ is homeomorphic to an open subset of $\mathbb{R}^k$. Will I get somewhere from here? I don't know.
Any hints or ideas would be very helpful. Thanks!
 A: An embedding is a smooth map $f:M\rightarrow N$ between manifolds $M$ and $N$ that is an immersion and also a homeomorphism onto its image $f(M)$, where this image has been given the subspace topology inherited from $N$.  You then should find what you're looking for in the Constant Rank and Immersion theorems of, for instance, $\S11$ of Tu.
A: As an exercise, I'll try to answer my own question.
First, as @g.s noticed, "$f$ is an embedding" means not only that $f:M\to f(M)$ is a homeomorphism, but also that $f$ is an immersion, which I've missed. Now, for one direction:
$(\Rightarrow)$ Let $k:=\dim M$. Since $f$ is an immersion, $\text{rank} f_{*_{p}}=k$ for all $p\in M$. By the constant rank theorem, there are charts $(U,X)$ at $p\in M$ and $(W,Y)$ at $f(p)\in N$, such that:
$$Y\circ f\circ X^{-1}(a_1,...,a_k)=(a_1,...,a_k,0,...,0)$$
Therefore, $Y(f(U))\subset Y(W)\cap(\mathbb{R}^k\times\{0\}^{n-k})^{(*)}$. Since $f$ is a homeomorphism, $f(U)=W'\cap f(M)$ for some open $W'\subset N$. Besides, $Y$ is also a homeomorphism and $U$ is homeomorphic to an open subset of $\mathbb{R}^k$, therefore $Y(f(U))$ is also homeomorphic to an open subset of $\mathbb{R}^k$. From $(*)$ we may conclude that $Y(f(U))=Y(W'')\cap (\mathbb{R}^k\times\{0\}^{n-k})$ for some open $W''\subset N$. So, Defining $V:=W'\cap W''$, we get $Y(V\cap f(U))=Y(V)\cap(\mathbb{R}^k\times\{0\}^{n-k})$.
