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Reparametrization of Curves ... Tapp, Section 4, Ch. 1 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
I am reading Kristopher Tapp's book: Differential Geometry of Curves and Surfaces ... and am currently focused on Chapter 1: Curves ... and in particular on Section 4: Reparametrization ...

I need help in order to fully understand the proof of Proposition 1.25 ... ...

Proposition 1.25 and its proof read as follows:



Tapp - 1 - Proposition 1.25 ... PART 1 ... .png
Tapp - 2 - Proposition 1.25 ... PART 2 ... .png



My questions are as follows:


Question 1

In the above text from Tapp we read the following:

" ... ... By the fundamental theorem of calculus, \(\displaystyle s'(t) = \mid \gamma ' (t) \mid \neq 0\), from which it can be seen that \(\displaystyle s\) is a smooth bijection onto \(\displaystyle \tilde{I}\) with nowhere-vanishing derivative. ... ... "


My question is as follows:

How do we know that \(\displaystyle s\) is a smooth bijection onto \(\displaystyle \tilde{I}\) ... how would we rigorously demonstrate this ... ... ?



Question 2

In the above text from Tapp we read the following:

" ... ... Therefore, \(\displaystyle s\) has an inverse function \(\displaystyle \phi \ : \ \tilde{I} \to I\), which is also a smooth bijection with nowhere-vanishing derivative. ... ... "


My question is as follows:

How do we know that \(\displaystyle \phi\) is a smooth bijection with nowhere-vanishing derivative. ... ...

Presumably it is because the inverse of a smooth bijection is also a smooth bijection ... is that correct ...?



Hope someone can help ...

Peter
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
Hi, Peter !

Since $s'(t) = \lvert \gamma'(t)\rvert > 0$, then $s$ is strictly increasing on $I$. Therefore, $s$ is one-to-one. By definition $\tilde{I} = s(I)$, so $s : I \to \tilde{I}$ is onto. Being one-to-one and onto, $s$ is a bijection from $I$ onto $\tilde{I}$. From calculus, we know that $\phi$ is differentiable with derivative $\phi'(u) = \dfrac{1}{s'(\phi(u))}$. Moreover, $\phi$ is a bijection since $s$ is the inverse of $\phi$. What definition exactly is being used here for smoothness?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,886
Hobart, Tasmania
Hi, Peter !

Since $s'(t) = \lvert \gamma'(t)\rvert > 0$, then $s$ is strictly increasing on $I$. Therefore, $s$ is one-to-one. By definition $\tilde{I} = s(I)$, so $s : I \to \tilde{I}$ is onto. Being one-to-one and onto, $s$ is a bijection from $I$ onto $\tilde{I}$. From calculus, we know that $\phi$ is differentiable with derivative $\phi'(u) = \dfrac{1}{s'(\phi(u))}$. Moreover, $\phi$ is a bijection since $s$ is the inverse of $\phi$. What definition exactly is being used here for smoothness?


Thanks for the help Euge ...

Very much appreciate your help ...

Definition of smoothness is given in text shown below ...



Tapp - Definition of Smoothness .png


Peter