Can diffeomorphism $i^{-1}:\Omega' \to \Omega$ be extended to a smooth function with an open domain set. Suppose I have an open subset $\Omega\subseteq {\mathbb R}^n$ and a smooth (i.e. in $C^\infty$) injective mapping $i: \Omega\to {\mathbb R}^N$ where $N\geq n$. Does this mean that there is a smooth mapping $f: A\to \Omega$ such that $A\supseteq i(\Omega)$, $A$ is open in ${\mathbb R}^N$ and $f|_{i(\Omega)} = i^{-1}$.
It can be shown that $\Omega$ and $i(\Omega)$ are are diffeomorphic as smooth manifolds. Argument: there is a single chart on $\Omega$, $(\Omega, {\rm id})$ and a single chart on $i(\Omega)$, i.e. $(i(\Omega), i^{-1})$. It is easy to see that $i: \Omega\to i(\Omega)$ is a $C^\infty$ mapping between smooth manifolds $\Omega$ and $i(\Omega)$, and its inverse $i^{-1}: i(\Omega)\to \Omega$ is also $C^\infty$ mapping.
Thus, $\Omega$ and $i(\Omega)$ are diffeomorphic. Does this mean that there is a smooth mapping $f: A\to \Omega$ such that $A\supseteq i(\Omega)$, $A$ is open in ${\mathbb R}^N$ and $f|_{i(\Omega)} = i^{-1}$.
 A: While it is true you can give the image $i(\Omega)$ the structure of a smooth manifold such that $\Omega$ and $i(\Omega)$ are diffeomorphic, it doesn't help you because this structure won't be necessarily "related" to the structure of the ambient space $\mathbb{R}^N$ in which $i(\Omega)$ sits. When you use the map $i$ to define a chart on $i(\Omega)$, you also force the topology on $i(\Omega)$ to be "identical" to the topology on $\Omega$ and this not necessarily the subspace topology $i(\Omega)$ inherits from $\mathbb{R}^N$.
For example, when $\Omega = (-\pi,\pi)$ and $i \colon \Omega \rightarrow \mathbb{R}^2$ is given by $i(t) = (\sin 2t, \sin t)$ the image of $\Omega$ looks like a figure-eight curve:

In this case, the inverse map $i^{-1} \colon i(\Omega) \rightarrow \Omega$ isn't continuous (as $i(\Omega)$ is compact while $\Omega$ is not) so you definitely don't have a smooth (or continuous) map $f \colon A \rightarrow \Omega$ such that $f|_{i(\Omega)} = i^{-1}$. However, if $i$ is an injective embedding then $i(\Omega)$ is an embedded submanifold and in this case you can find an open set $A$ containing $i(\Omega)$ and a smooth map $f \colon A \rightarrow \Omega$ such that $f|_{i(\Omega)} = i^{-1}$. This follows from the fact that $i^{-1} \colon i(\Omega) \rightarrow \Omega$ is a smooth map defined on an embedded submanifold and such a map can always be extended to a smooth map defined on an open subset of the ambient space which contains the embedded submanifold (see problem 5.17 in Lee's Introduction to Smooth Manifolds).
A: You need some additional assumptions on the set $\Omega$ at least. Consider the case when $\Omega$ is not connected, then if $A$ is connected, $i^{-1}$ can't be extended even to the continuous mapping of $A$.
