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, it is confirmed that the example is an error. 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 (extension is hopefully possible because "Most of the concepts introduced for a manifold extend word for word to a manifold with boundary...").

It was pointed out by quarague in this question's comments that another condition that might make the example true is to assume the image, which is given to be 1-manifold with boundary, is additionally assumed to have non-empty boundary. As it turns out, this might actually imply $c$ injective.

  1. Is it true that $c$ would be injective?

  2. If yes to (1), then why?

  3. If no to (1), then is the example still true even though $c$ is not injective?

  4. If no to (1) and yes to (3), then why?

  5. If no to (1) and no to (3), then why?

Note: Maybe this is now relevant.


Yes. The only $1$-manifolds, up to diffeomorphism, are $S^1, \mathbb R$, $[0, 1]$ and $[0, 1)$.

Fact. For $X$ $1$-dimensional and $Y$ $1$-dimensional, with $Y$ not diffeomorphic to $S^1$, a smooth immersion $c : X \to Y$ is injective.

Proof. Take a point $t \in X$ such that $c(t)$ equals some $c(s)$ with $s \neq t$. Because $c$ is an immersion, $c$ is approximately linear around $t$, so there exists $s \in X$ and a segment $[t, s] \subset X$ such that $c(s) = c(t)$ and such that $c$ is injective on $[t, s)$. Here, by segment, I mean a diffeomorphism on its image $\phi : [p, q] \to X$ with $\phi(p) = t$, $\phi(q) = s$. Now $c \circ \phi$ is a smooth immersion $[p, q] \to \mathbb R$ with $\phi(p) = \phi(q)$. By the mean value theorem, its derivative vanishes somewhere. A contradiction. $\square$

To prove that classification, here's a hint: consider a maximal chart $(0,1) \to X$. Use maximality to prove that its image misses at most $2$ points of $X$.

  • $\begingroup$ Thanks. I think you can answer here too. Is there a way to do this without classification theorem because classification theorem isn't given in this book? $\endgroup$ – user636532 Jun 26 '19 at 8:53
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    $\begingroup$ Yes, in the sense that the only part of the classification that we need for the original question, is that a $1$-manifold with nonempty boundary can be embedded in $\mathbb R$. (The part you cite in the other answer.) $\endgroup$ – Bart Michels Jun 26 '19 at 8:56
  • $\begingroup$ Nice. Something I'll check out later. Thanks! $\endgroup$ – user636532 Jun 26 '19 at 8:57
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    $\begingroup$ I rejected your edit suggestion: we only need $Y$ to be not diffeomorphic to $S^1$, because we only need $Y$ to be embeddable in $\mathbb R$. $\endgroup$ – Bart Michels Jun 26 '19 at 10:03
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    $\begingroup$ It was a typo, thanks. $\endgroup$ – Bart Michels Jun 26 '19 at 10:04

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