Diffeomorphisms between subsets of $\mathbb{R}^n$ I would like to know if the following statement and the following argument are correct:

Statement: Let $f:U\to A$ be a diffeomorphism where $U$ is an open subset of $\mathbb{R}^n$ and $A$ is a subset of $\mathbb{R}^n$. Then $A$ is also open in $\mathbb{R}^n$.

For clarity: if $A$ is a subset of $\mathbb{R}^n$ and $f:A \to \mathbb{R}^m$ is a map, I say that $f$ is smooth on $A$ if for every $y\in A$ there is an open subset $W$ of $\mathbb{R}^n$ containing $y$ and a smooth map $\eta:W\to \mathbb{R}^n$ such that $f=\eta$ on $A \cap W$.
If $A$ and $B$ are subsets of $\mathbb{R}^n$, a map $f:A\to B$ is a diffeomorphism if it is bijective and both $f$ and $f^{-1}$ are smooth in the above sense.

Argument: Let $y_0\in A$. I want to show that there is an open subset V of $\mathbb{R}^n$ such that $y_0 \in V \subseteq A$. Since $f^{-1}$ is smooth on $A$, then there is an open subset $W$ of $\mathbb{R}^n$ containing $y_0$ and a smooth map $\eta:W\to \mathbb{R}^n$ such that $f^{-1}=\eta$ on $A \cap W$. Since $f$ is continuous, we have that $f^{-1}(W)$ is open in $U$ and thus in $\mathbb{R}^n$. Now consider $f$ as a function from $f^{-1}(W)$ to $W$, which is still smooth. We have that $\eta \circ f=Id_{f^{-1}(W)}$. From the chain rule we have $D\eta(f(x)\cdot Df(x)=I_n$, so the matrix $Df(x)$ is invertible for every $x$ in $f^{-1}(W)$. By the inverse function theorem the map $f:f^{-1}(W)\to \mathbb{R}^n$ is open. Thus $V=f(f^{-1}(W))=W \cap A$ is open in $\mathbb{R}^n$, and $y_0 \in V \subseteq A$ as desired.

 A: Your argument is correct. I only have a minor suggestion (which is perhaps a bit nitpicking). You do not explicitly define what is means that a map with range $A \subset \mathbb R^n$ is smooth if $A$ is not known to be open. The solution is of course to regard $f$ as a map from $U$ to $\mathbb R^n$. Thus your statement could be reformulated as follows: Let $f : U \to \mathbb R^n$ be a smooth injective map and let $f^{-1} : f(U) \to U$ be smooth.
As Hudson Lima noticed in his comment, you show that for each $x_0 \in U$ there exist open neigborhoods $W'$ of $x_0$ in $U$ and $W$ of $f(x_0)$ in $\mathbb R^n$ and a smooth map $\eta : W \to W'$ such that $f(W') \subset W$ and $\eta \circ f = id$ on $W'$. This map $\eta$ is a local left inverse for $f$. It implies that $Df(x_0) : \mathbb R^n  \to \mathbb R^n$ is an injective linear map, hence invertible.
If you start with a smooth injection $f : U \to \mathbb R^m$, then you see that necessarily $n \le m$.
Perhaps you know that there is much stronger result known as invariance of domain: If $f : U \to \mathbb R^n$ is a continuos injection, then $f(U)$ is open and $f$ is a homeomorphism between $U$ and $f(U)$. See  https://en.wikipedia.org/wiki/Invariance_of_domain.
