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Context: 1. Are manifold subsets submanifolds? 2. Can manifold subsets always be made into submanifolds? 3. Why is the inclusion from a submanifold smooth?

Let $A,B$ be topological spaces with $A \subseteq B$ and $A$ a topological subspace of $B$. Suppose $A$ and $B$ become smooth manifolds $(A,\mathscr A)$ and $(B,\mathscr B)$ with respectively with dimensions $a$ and $b$.

  1. What's an example where the inclusion map $\iota: (A,\mathscr A) \to (B,\mathscr B)$ is smooth and a topological embedding but not an immersion?

There are a lot of examples of smooth immersions that are not topological embeddings (and thus not smooth embeddings) like this. There are also examples of smooth topological embeddings that are not immersions (and thus, again, not smooth embeddings) like this. In some of the questions linked above, there were examples that where $\iota$ wasn't smooth or even continuous. The purpose of this question is to ask specifically about the inclusion map and the case that the inclusion map is smooth. If such examples exists, then this tells me there's nothing particularly different about the inclusion map.

  1. What's an example where the inclusion map $\iota: (A,\mathscr A) \to (B,\mathscr B)$ is smooth but not a topological embedding?

Just checking my understanding. If there are no such examples, then (1) could simply ask "smooth but not immersion".

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    "Let $A,B$ be topological spaces with $A \subseteq B$ and $A$ a topological subspace of $B$." Then $\iota$ is automatically a topological embedding. Or do you mean that $\iota$ should embed $A$ as a topological submanifold of $B$ (that is, $A$ has to be a topological submanifold of $B$)? – Paul Frost Oct 17 '19 at 12:26
  • @PaulFrost I meant topological-subspace-embedding. Thanks. That answers Question 2 and now... –  Oct 18 '19 at 03:48

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Question 1:

Let $A = \mathbb R \times \{ 0 \}$ and $B = \mathbb R^2$. Then $A$ is a a topological submanifold of $B$. We give $B$ its standard smooth structure (with single chart atlas $id : B \to \mathbb R^2$) and $A$ the smooth structure with single chart atlas $f : A \to \mathbb R, f(x,0) = \sqrt[3]{x}$. Then $\phi = id \circ f^{-1} : \mathbb R \to \mathbb R^2$ is smooth since $\phi(x) = (x^3,0)$. Thus $\iota$ is smooth. However, it is not immersion because $T_{(0,0)} \iota$ is the zero map.

Paul Frost
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  • ...This answers Question 1. Thanks! –  Oct 18 '19 at 03:48
  • Oh this is based on a previous example of an injective but non-inclusion map but made it really an inclusion map by changing Its domain, $\mathbb R$, into its injective image, $\mathbb R \times {0}$, thus showing there's nothing particularly different for an inclusion map from any other injective map? –  Oct 18 '19 at 08:07
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    We can even simplify the example. Let $A = B = \mathbb R$. Then $\iota = id$ is a topological embedding of $1$-manifolds, but it is no an immersion if we give $B$ ist usual smooth structure and $A$ the smooth structure with single chart atlas $f : A \to B, f(x) = \sqrt[3]{x}$. – Paul Frost Oct 18 '19 at 23:00
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    That is, the core of the example is that there exist topological manifolds having more than one smooth structure. The point is that there are homeomorphisms which are no diffeomorphisms. Observe, however, that in the above example the two smooth manifolds are still diffeomorphic. A diffeomorphisms $\phi : A \to B$ is given by $\phi(x) = x^3$. – Paul Frost Oct 18 '19 at 23:12
  • That was an exercise in An Introduction to Manifolds by Loring W. Tu, but we didn't have immersions until about 2 or 3 sections later. Thanks! –  Oct 19 '19 at 02:16
  • Oh wait I re-read one of Eric Wofsey's comments. It's actually the same example as yours and as the one here and thus my question is kind of a duplicate? –  Oct 22 '19 at 11:34