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My book is An Introduction to Manifolds by Loring W. Tu. Pictured below is the last example from Section 22, Manifolds with Boundary.

enter image description here

I have been trying to wrap my head around this for about 2 hours (3.5 hours, if you include the 1.5 hours spent on the other question).

An alternative way I approach the example: classification of smooth 1-manifolds with boundary

  1. Does the classification of smooth 1-manifolds with boundary imply $C = c[a,b]$ and $[a,b]$ are diffeomorphic to $[0,1]$ and thus diffeomorphic to each other and thus $\partial C$ is diffeomorphic to $\partial [a,b] = \{a,b\}$, and $C^o$ is diffeomorphic to $(a,b)$?

  2. Must $c$ be an embedding and in particular $c$ is injective?

Update: The classification theorem assumes nonempty boundary. Let's do so as well otherwise the example is wrong. Also see here for the same question but not specifically to do with the classification theorem.

1 Answers1

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You are right that the classification does not imply that $C$ and $[a, b]$ are diffeomorphic. It may very well be that $C$ is diffeomorphic to a circle, and that $c$ winds the interval $[a, b]$ 3.5 times around it. (The minute hand of the clock was giving you a hint.)

But assuming that $C$ has nonempty boundary, it follows indeed that the immersion $c$ is injective (i.e. is an embedding), as shown in a linked question: Is a smooth immersion $c: [a,b] \to M$ injective if its image is a 1-manifold with non-empty boundary?

In particular, the boundary of $C$ has cardinality $2$ and equals $\{ c(a), c(b) \}$.

Bart Michels
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  • LOL. Thanks –  Jun 26 '19 at 09:24
  • punctured dusk, actually, do you or do you not assume $C$ is a 1-submanifold with boundary, namely with non-empty boundary of $M$? I'm just checking to see if this still makes sense if $C$ has some arbitrary differentiable structure that isn't really related to $M$ or something. –  Aug 08 '19 at 16:17
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    I do assume $C$ is a submanifold, if only to define the differentiable structure of $C$. I can't otherwise see a canonical way to give $C$ a differentiable structure. Then again, I think that the boundary of a $C^1$-variety can be defined topologically. If this is true, then a posteriori it doesn't matter which $C^1$-structure we give $C$, as long as we don't change the topology. – Bart Michels Aug 09 '19 at 10:14
  • punctured dusk, thanks! You can answer here if you want. –  Aug 14 '19 at 03:59
  • Bart Michels, is $c$ a local diffeomorphism onto image? See OP's 2 other posts: Surjective immersions and local diffeomorphism onto image – BCLC Apr 28 '21 at 03:55
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    @BCLC Yes, an immersion (that is, a differentiable map with everywhere injective differential) is locally a diffeomorphism onto (local) image. – Bart Michels Apr 28 '21 at 09:59
  • i know immersion is equivalent to local embedding aka local-(diffeomorphism onto image), but i'm wondering: is this $c$ in particular is a (local difeomorphism)-onto image? the latter concept of LDOI (as in (LD)-OI) is stronger than immersion (as in L-(DOI)). it's proven that an immersion between manifolds whose image is not merely an immersed submanifold but actually a regular/embedded submanifold is a (local diffeomorphism)-onto image. i guess the previous sentence would apply to manifolds with boundary too. – BCLC Apr 29 '21 at 07:14
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    @BCLC In any case, in the one-dimensional case the image of a smooth immersion is a smooth submanifold (with or without boundary) of the target. And then there is no ambiguity in LDOI. – Bart Michels Apr 29 '21 at 07:24
  • Bart Michels, do you mean that in 1 - the condition that the image of $c$ is reg/emb submanifold with boundary (with non-empty boundary), the only assumption is the non-empty boundary because the image is already reg/emb submanifold with boundary and 2 - if so, then this is because the domain of $c$ is 1-dimensional? – BCLC Apr 29 '21 at 07:34
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    @BCLC Yes, the fact that the map is an immersion implies that the image is a smooth submanifold with boundary, and this is because both domain and target manifold are 1-dimensional. (It would not be always true if only the domain is 1-dim.) So that, as you say, the only extra condition you need is that the image has nonempty boundary. – Bart Michels Apr 29 '21 at 09:32
  • thanks Bart Michels! – BCLC Apr 29 '21 at 12:04