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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?

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    You can even endow $X$ with a (unique) smooth structure that makes the map $f:X\rightarrow N$ a diffeomorphism. First pull back the topology of $N$: declare open subsets of $X$ to be exactly the preimages under $f$ of open subsets of $N$. Then pull back the smooth structure: if ${(U_{i},\phi_{i}}$ is the smooth structure on $N$, then ${(f^{-1}(U_{i}),\phi_{i}\circ f)}$ is a smooth structure on $X$, and $f:X\rightarrow N$ becomes a diffeomorphism. – studiosus Jul 31 '19 at 13:37
  • "You can even..." is an additional comment to an answer of "Yes" ? @studiosus –  Jul 31 '19 at 13:39
  • @studiosus By the way, your comment assumes or doesn't assume $f$ is bijective? –  Jul 31 '19 at 13:41
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    @studiosus but you are putting smooth structure on a set. Is the topology induced by your smooth structure the same as the subspace topology on $X$? –  Jul 31 '19 at 13:49
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    what do you mean when you ask if $X$ is a manifold? (1) There is a manifold that has $X$ as its underlying set. (2) There is a manifold that has $X$ with the subspace topology as its topological space. (3) $X$ is a submanifold of $M$ (4) none of the above. –  Jul 31 '19 at 13:58
  • @ljr Oh, what a great question. Thanks! Um, (3) I guess based on the context. I'll edit my question. –  Aug 01 '19 at 03:06
  • Also, @ljr I think studiosus implicitly answers "Yes" without proof to the question and then additionally comments that we can kind of do the converse...Or maybe studiosus is actually thinking I mean "Does there exist a smooth structure on $X$ for $f$ given as a diffeomorphism in this sense?", and then answers "Yes" implicitly and then answers converse...Not sure. I think my question was wrong based on your second comment. –  Aug 01 '19 at 03:08
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    @SeleneAuckland To be clear, all I'm saying is that if you have a set $X$, a manifold $N$ and a bijection $f:X\rightarrow N$, then you can give $X$ a smooth manifold structure. And even more, you can do this in such a way that $X$ becomes diffeomorphic to $N$ through the map $f$. So this answers your original question, namely whether $X$ is a smooth manifold. – studiosus Aug 01 '19 at 08:15
  • @studiosus Ok, that tells me my question was wrong in conflicting what what I intended to ask. Thanks! –  Aug 01 '19 at 08:46
  • @studiosus in re your 1st comment, is it really unique? or just unique up to diffeomorphism like as described in this comment? (or maybe there isn't a difference...) – BCLC Apr 28 '21 at 02:38

1 Answers1

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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.

Claire
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