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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.

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In this question, Prof Jack Lee says that the example is incorrect as stated. Besides a few nitpicks such as $c(p)$ instead of $p$ or $c:[a,b] \to M$ versus restricted range $c$ given by $\tilde{c}:[a,b] \to C$, a counterexample given was $c:[0,2\pi] \to \mathbb R^2$, $c(t)=(\cos t, \sin t)$ (I guess you can do $e^{it}$ and $\mathbb C$ if you like). However, the example is true if $c$ were injective by Exercise 11.5 extended from manifolds to manifolds with boundary (Exercise 11.5 refers to $N$ and $M$ as manifolds with boundary without boundary, but we can hopefully extend Exercise 11.5 for $N$ and $M$ as manifolds with boundary because "Most of the concepts introduced for a manifold extend word for word to a manifold with boundary...").

As alternatives to $c$ being injective:

  1. What if we assume $c$ is not injective but is a local diffeomorphism?

  2. What if we assume $c$ is not injective but $\tilde{c}$ is a local diffeomorphism?

  3. What if we assume $c$ is not injective but $c$ is a local diffeomorphism onto its image? (Ignore if this is the same question as (2)...which it might not be based on this. I indirectly ask about this here)

  4. What are other conditions we can have on $c$ to make the example true without assuming $c$ injective?

    • Update A: Based on quarague's comments below, it seems one condition to make the example true is to assume the image has non-empty boundary. Then this would make $c$ injective and the example true. This is technically beyond the letter of the scope of the example, but definitely in the spirit. I'll ask about this in another question. Maybe this was actually the critical error, but I'm curious as to how this might be true for non-injective (perhaps we could somehow have non-injective but with non-empty boundary).

      • Update B: Asked here.

Update 1: Based on "Most of the concepts introduced for a manifold extend word for word to a manifold with boundary..." and on this, I think $c$ is indeed a local diffeomorphism onto image if and only if $C$ were a 1-submanifold with boundary of $M$ (as opposed to simply being an arbitrary 1-manifold with boundary that happens to be a subset of $M$. I ask about this here). Therefore, I think alternatives (2) and (3) are ruled out.

Update 2: As for alternative (1), if $c$ were assumed to be a local diffeomorphism, then $C$ becomes not only some 1-manifold with boundary that happens to be a subset of $M$ but actually a regular/an embedded 1-submanifold with boundary of $M$, in particular with codimension zero since $C$ would be open in $M$ since local diffeomorphisms are open maps. Hence, $\dim M = 1$ (assuming $M$ has a dimension since in this book not all manifolds with boundary have dimensions). I don't think $M$ is given to be a manifold with non-empty boundary, so if $M$ were a manifold ($\partial M = \emptyset$), then I guess $C$ would have to be a manifold too ($\partial C = \emptyset$). If $M$ were a manifold with non-empty boundary, then while we have that $C$ is a manifold with boundary, I don't see any guarantee that $C$ has non-empty boundary.

Update 3: It appears injectivity of $c$ is equivalent to $\partial C \ne \emptyset$ based on the question in Update B, and thus the answer to this question is

This is an issue of "with boundary" meaning "with non-empty boundary" (see quarague's comments) and that of injectivity of $c$ is equivalent to $\partial C \ne \emptyset$ (see punctured dusk's answers here and here).

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    $\begingroup$ The first sentence states that the image of $c$ is a $1$-d manifold with boundary. This is not the case in your counterexample. This condition also implies that $c$ is injective. $\endgroup$ – quarague Jun 21 at 7:25
  • $\begingroup$ @quarague Manifold with boundary includes manifold as a manifold with boundary with empty boundary. Did you mean the statement is true if we have non-empty boundary like here because non-empty boundary implies $c$ injective? $\endgroup$ – Selene Auckland Jun 21 at 7:31
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    $\begingroup$ This is about interpreting 'manifold with boundary'. If you think this includes manifolds with empty boundary (which often makes sense) than the example in Tu's book doesn't work. It makes more sense to me to interpret it here as manifold with nonempty boundary in which case I think the example is correct. Nonempty boundary + immersion would imply injective in this example. $\endgroup$ – quarague Jun 21 at 8:40
  • $\begingroup$ @quarague Ok I'll think about that or ask that in another question. Thanks! $\endgroup$ – Selene Auckland Jun 21 at 8:52
  • $\begingroup$ @quarague Update: Asked here. $\endgroup$ – Selene Auckland Jun 21 at 9:35

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