Is being diffeomorphic to a manifold equivalent to being a manifold? Obviously manifolds are diffeomorphic to manifolds (namely themselves by the identity map). Is the converse true with diffeomorphism in this sense?
To be explicit: Let $M$ and $N$ be (smooth) manifolds. If need be, then you may give them dimension (Note: In some textbooks, not all manifolds have dimension). Let $X$ be a subset of $M$. Is $X$ a (regular/an embedded) submanifold of $M$ if there exists a map $f:X \to f(X)=N$ that is a diffeomorphism in this sense?


*

*Edit: I previously asked if $X$ was a manifold but based on ljr's comment and I guess based on this question and this question, I guess asking for $X$ to be a manifold is not a very good question. 


We have that:


*

*such $f$ is bijective

*such $f$ is smooth in this sense: For each $p \in X$, there exists a neighborhood $U_p$ of $p$ in $M$ and a smooth map $g: U_p \to N$ such that the restrictions $g|_{U_p \cap X}: U_p \cap X \to N$ and $f|_{U_p \cap X}: U_p \cap X \to N$ agree on $U_p \cap X$: $g|_{U_p \cap X} = f|_{U_p \cap X}$.

*the inverse of such $f$, $f^{-1}$, is smooth in this sense: For each $q=F(p) \in N$, with $p \in X$, there exists a neighborhood $V_q$ of $q$ in $N$ and a smooth map $h: V_q \to M$ such that the restrictions $h|_{V_q \cap N = V_q}: V_q \to M$ and $f^{-1}|_{V_q}: V_q \to X$ agree on $V_q$: $h|_{V_q} = f^{-1}|_{V_q}$.
So far I've thought of extending $h$ to $\tilde h: N \to M$ (in whatever extension possible given $h$ might not have compact support), of $\tilde h(V_q)=h(V_q)=f^{-1}(V_q)$ possibly being a subset of $X$ or something and of this.
I don't know if the above counts as effort towards answering the question, but if the above doesn't, then may you just please provide a link proving or providing a counterexample and then I'll just work out the details myself (I would think of the justification or counterargument after I know what the answer is)?
Context: Are immersed submanifolds something like local manifolds the same way manifolds are locally Euclidean?
 A: I think it is true that $X$ is an embedded submanifold. 
In the following all inclusions will be denoted by $i$.
Let $J=i\circ f^{-1}: N\to M$. Then by Theorem 11.13 of Tu's Introduction to Smooth Manifolds (2nd edition) it is enough to show that $J$ is an immersion and a homeomorphism onto its image.


*

*Functions defined on arbitrary subsets on manifolds which are smooth in the news sense are continous (when the subsets are given the subspace topology) so in particular $f$, $f^{-1}$ are continous and hence $J$ is a homeomorphism onto its image.

*From 3. it follows that $J$ is smooth in the old sense since with the notation above $h=J_{|V_q}$ and being smooth is a local property. 

*For each $q=f(p)\in N$ let $U_p$ and $g$ as in 2. and set $W_q=f(U_P\cap X)$ which is an open neighbourhood of $q$. Let $J':W_q\to U_p$ be the rectriction of $J$. Then $g\circ J'=i$ and as $i$ is an immersion so is $J'$. Then $J_{|V_q}=i\circ J'$ is an immersion as a composition of immersions. This shows that locally $J$ is an immersion and hence an immersion.
