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  1. If $B$ is a manifold, and $A\subseteq B$ is a regular submanifold, then the inclusion map $i:A\to B$ is an embedding and thus smooth.

That $i$ is an embedding seems a bit strong. Is there another way to do this?

  1. I seem to recall so far in learning differential geometry that inclusions are smooth specifically when doing compositions, but I can't find it upon looking through my notes. In general, for any manifold subset $A$ (a subset that is also a manifold but not necessarily a regular submanifold or immersed submanifold or neat submanifold, etc (I think not all irregular submanifolds are manifolds anyway)) of a manifold $B$, I know $i:A \to B$ is continuous (assuming subspace topology), but when is $i:A \to B$ smooth?

    • Upon second look of my notes, I found that it was true for the 'inclusion' $i_b: A \to A \times B, i_b(a)=a \times b$ for fixed $b$. That's different from what I usually think of inclusion and instead more like this 'inclusion'.

    • Upon third look, I think many of those subset manifolds $A$ were open in $B$. Since open implies regular submanifold, all those compositions were safe, but we didn't have submanifolds then. I ask about this in another question.

Thanks in advance!

user26857
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Ekhin Taylor R. Wilson
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    The inclusion is the restriction of the identity map (which is smooth) to $A$. Do you claim that it may not be smooth? That's odd. – Yanko Jan 18 '19 at 12:09
  • @Yanko I knew it had something to do with identity's being smooth! The thing is I saw a proof that said an inclusion was smooth because it was from a regular submanifold $A$, of a manifold $B$, to $B$. I was surprised. I thought identity and inclusion were always smooth (under subspace topology or some other obvious assumption) So is $i: A \to B$ for $A$ and $B$ smooth manifolds and $A \subseteq B$ always smooth whether $A$ is regular, immersed, neat or not even a submanifold? – Ekhin Taylor R. Wilson Jan 19 '19 at 17:08
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    I think that the inclusion is smooth if and only if $A$ is an embedded submanifold, as strong as it may seem. Try looking in the book of Warner on differentiable manifolds. – Giuseppe Negro Jan 19 '19 at 17:31
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    @Yanko: https://math.stackexchange.com/q/2296838/8157 This is a typical example of a subset of $\mathbb R^2$ that is a differentiable manifold but it is such that the inclusion is not smooth. I agree that these things are odd, though. – Giuseppe Negro Jan 19 '19 at 17:32
  • @GiuseppeNegro Okay I double checked. It's embedding if and only if embedded manifold. 'Inclusion is an embedding' is stronger than 'inclusion is smooth'? – Ekhin Taylor R. Wilson Jan 19 '19 at 17:48
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    "Inclusion is an embedding" surely implies "inclusion is smooth"; this is easy. In my previous comment I said that they are equivalent. I am not so sure anymore. You will have to search on books or on the net, unfortunately. – Giuseppe Negro Jan 19 '19 at 18:16
  • @GiuseppeNegro is my answer correct? – BCLC Apr 26 '21 at 03:40

1 Answers1

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The inclusion map is automatically an embedding if it is smooth unless...

  1. 1stly, talking about an inclusion map as smooth means both the objects in question are manifolds.

  2. The inclusion map is an embedding if and only if the domain is a regular/an embedded submanifold of the range. It's possible this isn't the case if the domain is given a manifold (or even topological!) structure s.t. the domain is not a regular/an embedded submanifold of the range. See here: Are manifold subsets submanifolds?

BCLC
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