Is being diffeomorphic to a manifold equivalent to being a manifold?

Question

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:

1. such $f$ is bijective

2. 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 $U_p \cap X$: $g|_{U_p \cap X} = f|_{U_p \cap X}$.

3. 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?

Answer 1

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.

1. 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.

2. 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.

3. 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.