If a curve $c$ can be reparametrized by arc length, then is $c$ regular? 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).
I refer to Section 2.1, Section 2.2, Volume 1 Section 8.6 (Part 1), Volume 1 Section 8.6 (Part 2) and this question: Does "$s'(t_0) = ||c'(t_0)||$" actually mean "$s'(t_0) \cong ||c'(t_0)||$" or "$\dot s(t_0) = ||c'(t_0)||$"?

Question: Proposition 2.3 says if $c$ is regular, then $s$ is a diffeomorphism. Conversely, If $s$ somehow is a diffeomorphism, then is $c$ regular?


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*Context: I'm wondering if Proposition 2.4 should instead start "A regular curve is parametrized..." To be clear, I think Proposition 2.4 says:


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Let $c$ be a (not sure if regular or not) parametrized curve. If $c:[a,b] \to M$, for real numbers $a$ and $b$, can be reparametrized by arc length, i.e. we can define a (regular I guess) curve $\gamma := c \circ t: [\alpha,  \beta] \to [a,b] \to M$ for some diffeomorphism $t$ and for some real numbers $\alpha$ and $\beta$, then $\gamma$, the reparametrized $c$, is also regular and has unit speed and $\alpha = 0.$ (Of course, we later show $\beta = l$.) Conversely, if $c:[a,b] \to M$, for real numbers $a$ and $b$ is such that $c$ has unit speed and $a=0$, ($b=l$ is not used here, but I have a feeling $b=l$ can substitute for $a=0$. Not sure.) then the parameter $t$ of $c=c(t)$, is actually arc length.



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*I also assume the condition, in Proposition 2.4, "is parametrized by arc length" (or "can be reparametrized by arc length", depending on your version) is equivalent to the conclusion of Proposition 2.3, which is the existence of (Note: Edited to add the word smooth) smooth $s^{-1}$. If this assumption is wrong, then the question in the title is different from the one in the body.


*Updates:


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*A. I forgot to mention in the previous paragraph that I think the the conclusion of Proposition 2.3 is that $s^{-1}$ exists and is smooth.

*B. I'm actually not sure $t$ is a diffeomorphism for the "is parametrized by arc length". It could be that $t$ is just smooth. But maybe $t$ is a diffeomorphism based on Section 2.1.

*C. I'm actually not sure $s^{-1}$ exists and is smooth is equivalent to "$s$ is a diffeomorphism". Actually, if $c$ is regular, then $s$ is smooth and has a smooth inverse too. If $s$ has an inverse that is smooth (or even just differentiable) and $s$ is smooth, then $c$ I guess is regular, as pointed out in answer below. However, If $s$ has an inverse that is smooth (or even just differentiable) but $s$ is not necessarily smooth, then I'm not sure $c$ is regular. Waiting for response in comments in answer below. I think we need $s$ at least differentiable, which I think follows from FTC, assuming the hypotheses of FTC are satisfied. (I actually ask about the use of FTC here.)



*More context:


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*Can speed be defined for a parametrized curve that is not regular/not an immersion?

*Does a curve parametrized by arc length have unit speed and its parameter starting at 0 even if not regular/not an immersion?
 A: I think one of the issues is that there is no explicit definition of what it means for a curve to be "parametrized by arc length".  The implicit definition use in the proof of 2.4 is that a curve is parametrized by arclength if it is given as   $\gamma(s)=c(t(s))$  for some regular curve $c$, with $t(s)$ the inverse of the arclength function $s(t)$ of $c$. Since Prop 2.3 guaratees that $t(s)$ exists only when $c$ is regular, this definition a priori only makes sense under the assumption that  $c$ is  regular (and of course then $\gamma$ can be proved to is regular as in the second part of Prop 2.4 (in fact $|\gamma'(s)|=1$ so $\gamma'(s)\neq 0$)).
Now your question can be interpreted as "can $t(s)$ exist for non-regular $c$"? 
The answer is that if you require $t$ to be differentiable (which is part of saying that $s$ is a diffeomorphism), then it follows that $c$ is regular, since $s(t(s))=s$ and differentiating $s'(t)t'(s)=1$, and by main thm of calculus $s'(t)=|c'(t)|$, so $|c'(t)||t'(s)|=1|$, so $c'(t)\neq 0$ (on the other hand, if you only require $t$ to be a continuous inverse for $s$ then you can find plenty of examples where $c$ is non-regular, and they can then be "reparametrized by a continuous change of parameter"; however the resulting $\gamma(s)$ has no reason to be continuous, much less smooth (but can be arranged to be smooth in some cases)). In summary, if $c$ is regular all the definitions make sense and and resulting $\gamma$ is smooth. If $c$ is not regular, $t(s)$ can not be smooth, but sometimes can still exists, however other things will usually break.
