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My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

Here are Section 2.1 and Section 2.2.

About Proposition 2.3, I believe what happens is that we show $s$ is an immersion (or equivalently a submersion or a local diffeomorphism, by this and by "Most of the concepts introduced for a manifold extend word for word to a manifold with boundary...", from Volume 1 Section 22) and conclude that its inverse is smooth. I believe we can show $s$ is an immersion without deducing $s$ is bijective.

Question 1: Is this right? Please verify.

  1. $||c'||: [a,b] \to (0,\infty)$ is a smooth map (See here) because $c$ is regular/an immersion.

  2. $||c'||$ is continuous by (1).

  3. $\dot s = ||c'||$ by fundamental theorem of calculus and (2) (See here for $\dot s$ vs $s'$).

  4. Use $t$ denote the standard coordinate (Volume 1 Section 8.6) on $[a,b]$ but $t_0$ to denote a point in $[a,b]$. Then $\dot s(t_0) = ||c'(t_0)||$, for each $t_0 \in [a,b]$.

  5. $s'(t_0)=\dot s(t_0) \frac{d}{dx}|_{s(t_0)}$, where $x$ is the standard coordinate (Volume 1 Section 8.6) on $[0,l]$, by Volume 1 Exercise 8.14, where $s'(t_0) := s_{*,t_0}[\frac{d}{dx}|_{s(t_0)}] \in T_{s(t_0)}[0,l]$ and where $s_{*,t_0}$ is $s_{*,t_0}: T_{t_0}[a,b] \to T_{s(t_0)}[0,l]$.

  6. $s'(t_0)=||c'(t_0)|| \frac{d}{dx}|_{s(t_0)}$, by (4) and (5).

  7. $s$ is an immersion if and only if $s_{*,t_0}$ is injective if and only if the kernel of $s_{*,t_0}$ consists only of $0_{t_0}$, the zero tangent vector of $T_{t_0}[a,b]$.

  8. The kernel of $s_{*,t_0}$ consists only of $0_{t_0}$, the zero tangent vector of $T_{t_0}[a,b]$, as shown below.

  9. Therefore, $s$ is an immersion by (7) and (8).

Proof of (8):

  • 8.1. Suppose there exists $X_{t_0} \in T_{t_0}[a,b]$ such that $s_{*,t_0}[X_{t_0}] = 0_{s(t_0)}$, the zero tangent vector of $T_{s(t_0)}[0,l]$.

  • 8.2. Such $X_{t_0}$ in (8.1) is $X_{t_0)} = a(t_0) \frac{d}{dt}|_{t_0}$ for some real number $a(t_0)$.

  • 8.3. $0_{s(t_0)} = s_{*,t_0}[X_{t_0}] = s_{*,t_0}[a(t_0) \frac{d}{dt}|_{t_0}] = a(t_0) s_{*,t_0}[\frac{d}{dt}|_{t_0}] = a(t_0)s'(t_0) = a(t_0)||c'(t_0)|| \frac{d}{dx}|_{s(t_0)}$, by (6), (8.1) and (8.2).

  • 8.4 $||c'(t_0)|| = 0$ if and only if $c'(t_0) = 0_{c(t_0)}$, the zero tangent vector in $T_{c(t_0)}M$.

  • 8.5. $c'(u_0)$ is not $0_{c(u_0)}$ for all $u_0 \in [a,b]$, which is the - definition of $c$ regular/an immersion.

  • 8.6. $||c'(u_0)|| \ne 0$ for all $u_0 \in [a,b]$ by (8.4) and (8.5).

  • 8.7. $0_{s(t_0)} = a(t_0) \frac{d}{dx}|_{s(t_0)}$ by (8.3) and (8.6).

  • 8.8. $a(t_0) = 0$ by (8.7) and by definition of $0_{s(t_0)} := 0 \frac{d}{dx}|_{s(t_0)}$

  • 8.9. $X_{t_0} = 0 \frac{d}{dt}|_{t_0} = 0_{t_0}$ by (8.1), (8.8) and by definition of $0_{t_0} := 0 \frac{d}{dt}|_{t_0}$.

Question 2: Do I use bijectiveness of s anywhere above?

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    What is even the point of this? You've turned something really simple into something really obtuse, and for what? You need that $s$ is a bijection because it is the inverse function to $s$ that is needed to reparametrize $c$ to obtain the arclength parametrization. – Paul Sinclair Jul 27 '19 at 20:11
  • @PaulSinclair Just wanted to clarify if bijectiveness is needed to prove immersiveness of $s$... –  Jul 30 '19 at 11:09
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    "Immersiveness" is not a particular useful notion for $s$. There is no reason to think of $s$ as a map between manifolds here. Instead, just consider it a function on an interval into the real line. It is true that $s$ can be considered to be an immersion into $\Bbb R$, or a diffeomorphism with its image, but that is because any two open intervals in $\Bbb R$ are diffeomorphic. The only thing required of a surjective smooth function between two open intervals to be a diffeomorphism is that the derivative is never $0$. – Paul Sinclair Jul 30 '19 at 13:52
  • @PaulSinclair Thanks. Re first comment: Good insight and intuition. –  Jul 31 '19 at 02:47
  • @PaulSinclair Re second comment: I know immersions are equivalent to local embeddings, and I guess local embeddings are "locally injective" (I never bothered to look up the precise definition for "locally injective"), but how are immersions injective? I think you meant to say $F_{,p}$ is injective if and only if $F$ is an immersion at p if and only if $\ker F_{,p} = 0_p$, but it sounds like you're saying "$F$ immersion implies $F$ injective". Please clarify –  Jul 31 '19 at 02:48
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    I apologize for that. I was thinking of an embedding. I deleted the comment because it was just plain wrong. – Paul Sinclair Jul 31 '19 at 21:19
  • @PaulSinclair You don't have to apologize. Thanks. You can answer if you want. –  Aug 01 '19 at 03:03

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