Showing a set is a smooth manifold by constructing a diffeomorphism to a smooth manifold I have two sets $M,N \subseteq \mathcal{E}$, which are subsets in a Euclidean space $\mathcal{E}$. I know, that $M$ is a smooth embedded manifold, and I have a diffeomorphism $\Psi:\mathcal{E} \rightarrow \mathcal{E}$ that maps $\mathcal{E}$ onto itself, while also $\Psi(M) = \Psi(N)$.
My question is, is there a clever/fast way of showing that $N$ must be also a smooth embedded manifold of the same dimension? Could I imply something about the tangent spaces of $N$ as well?
I understand that I could derive this from first principles, but is there maybe a theorem that would save the trouble?
Thanks
 A: When working  with embedded submanifolds i like to use the following theorems:

If $S$ is a smooth embedded submanifold of $Y$, then the inclusion $S\hookrightarrow Y$ is a smooth immersion and a homeomorphism onto its image.


If $f:X\to Y$ is a smooth immersion and a homeomorphism onto its image then $f(X)$ is an embedded submanifold of $Y$ and $f:X\to f(X)$ is a diffeomorphism.

Here $\mathrm{inc}\circ\Psi :M\to\mathcal E$ is a smooth immersion (as a composition of smooth immersions) and a homeomorphism onto its image (as the restriction of a homeomorphism). So its image $N$ is an embedded submanifold of $\mathcal E$ and $\Psi_{|M}:M\to N$ is a diffeomorphism. In particular the dimensions of $M$ and $N$ are the same and the tangent space $T_pM$ gets mapped to $T_{\Psi(p)}N$ isomorphically by $d_p\Psi$.
But also using only the definition shouldn't cause too much trouble. For completeness:
Let $q\in N$ and set $p=\Psi^{-1}(q)$. Then there is an open set $U\subseteq\mathcal{E}$ containing $p$ and a smooth function $h:U\to\mathbb R^{n-k}$ with $n=\dim\mathcal{E}$ and $k=\dim M$ such that $0$ is a regular value of $h$ and $U\cap M=h^{-1}(0)$. Then $h\circ \psi^{-1}:\Psi(U)\to\mathbb R^{n-k}$ is a smooth function defined in an open neighbourhood of $p$ having $0$ as a regular value with
$$(h\circ \Psi^{-1})^{-1}(0)=\Psi(h^{-1}(0))=\Psi(U\cap M)=\Psi(U)\cap N$$
which shows that $N$ is an embedded submanifold of $\mathcal{E}$.
