Proof of equivalence well defined function












1












$begingroup$


This is the definition:



Definition 1:



A function $f:Dsubseteq Rto R^n$ is said to be continuously differentiable of a $C^1$ function, if f is differentiable and the first derivative of f is continuous.



Definition 2:



A curve C in $R^n$ is smooth simple arc if C has a 1-1 $C^1$ parametrization of the form $f:[a,b] to R^n$, the points f(a) and f(b) are then called the end points of the arc$



Now comes the theorem:



Let C be a smooth simple arc in $R^n$ simply parametrised by a smooth 1-1 function, $f:[a,b]subseteq Rto R^n $ Then any smooth parametrisation $h:[c,d]subseteq Rto R^n$ of C is 1-1 and is equivalent to f.



Now as the definition of equivalent is not what I am wondering about so I am not going to post it. My question is simply with regards to how the proof starts:



It starts as follows:



Proof



The function $Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ and satisfies $h=fcircPhi$.



Now if are okey with this I understand the how the proves follow, however, I am getting stuck with the part where they state that $Phi$ is well defined.



Question



1) Why is $Phi$ well defined?



2) Why does it satisfy $h=fcircPhi$






proof-explanation curves parametrization







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
    $endgroup$
    – AnyAD
    Nov 29 '18 at 8:31










  • $begingroup$
    So if f is one to one it follows that $f^{-1}$ is well defined?
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:33


















1












$begingroup$


This is the definition:



Definition 1:



A function $f:Dsubseteq Rto R^n$ is said to be continuously differentiable of a $C^1$ function, if f is differentiable and the first derivative of f is continuous.



Definition 2:



A curve C in $R^n$ is smooth simple arc if C has a 1-1 $C^1$ parametrization of the form $f:[a,b] to R^n$, the points f(a) and f(b) are then called the end points of the arc$



Now comes the theorem:



Let C be a smooth simple arc in $R^n$ simply parametrised by a smooth 1-1 function, $f:[a,b]subseteq Rto R^n $ Then any smooth parametrisation $h:[c,d]subseteq Rto R^n$ of C is 1-1 and is equivalent to f.



Now as the definition of equivalent is not what I am wondering about so I am not going to post it. My question is simply with regards to how the proof starts:



It starts as follows:



Proof



The function $Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ and satisfies $h=fcircPhi$.



Now if are okey with this I understand the how the proves follow, however, I am getting stuck with the part where they state that $Phi$ is well defined.



Question



1) Why is $Phi$ well defined?



2) Why does it satisfy $h=fcircPhi$






proof-explanation curves parametrization







share|cite|improve this question











$endgroup$












  • $begingroup$
    It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
    $endgroup$
    – AnyAD
    Nov 29 '18 at 8:31










  • $begingroup$
    So if f is one to one it follows that $f^{-1}$ is well defined?
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:33
















1












1








1





$begingroup$


This is the definition:



Definition 1:



A function $f:Dsubseteq Rto R^n$ is said to be continuously differentiable of a $C^1$ function, if f is differentiable and the first derivative of f is continuous.



Definition 2:



A curve C in $R^n$ is smooth simple arc if C has a 1-1 $C^1$ parametrization of the form $f:[a,b] to R^n$, the points f(a) and f(b) are then called the end points of the arc$



Now comes the theorem:



Let C be a smooth simple arc in $R^n$ simply parametrised by a smooth 1-1 function, $f:[a,b]subseteq Rto R^n $ Then any smooth parametrisation $h:[c,d]subseteq Rto R^n$ of C is 1-1 and is equivalent to f.



Now as the definition of equivalent is not what I am wondering about so I am not going to post it. My question is simply with regards to how the proof starts:



It starts as follows:



Proof



The function $Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ and satisfies $h=fcircPhi$.



Now if are okey with this I understand the how the proves follow, however, I am getting stuck with the part where they state that $Phi$ is well defined.



Question



1) Why is $Phi$ well defined?



2) Why does it satisfy $h=fcircPhi$






proof-explanation curves parametrization







share|cite|improve this question











$endgroup$




This is the definition:



Definition 1:



A function $f:Dsubseteq Rto R^n$ is said to be continuously differentiable of a $C^1$ function, if f is differentiable and the first derivative of f is continuous.



Definition 2:



A curve C in $R^n$ is smooth simple arc if C has a 1-1 $C^1$ parametrization of the form $f:[a,b] to R^n$, the points f(a) and f(b) are then called the end points of the arc$



Now comes the theorem:



Let C be a smooth simple arc in $R^n$ simply parametrised by a smooth 1-1 function, $f:[a,b]subseteq Rto R^n $ Then any smooth parametrisation $h:[c,d]subseteq Rto R^n$ of C is 1-1 and is equivalent to f.



Now as the definition of equivalent is not what I am wondering about so I am not going to post it. My question is simply with regards to how the proof starts:



It starts as follows:



Proof



The function $Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ and satisfies $h=fcircPhi$.



Now if are okey with this I understand the how the proves follow, however, I am getting stuck with the part where they state that $Phi$ is well defined.



Question



1) Why is $Phi$ well defined?



2) Why does it satisfy $h=fcircPhi$






proof-explanation curves parametrization




proof-explanation curves parametrization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 8:38









José Carlos Santos

159k22126231




159k22126231










asked Nov 29 '18 at 8:15









ALEXANDERALEXANDER

8951921




8951921












  • $begingroup$
    It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
    $endgroup$
    – AnyAD
    Nov 29 '18 at 8:31










  • $begingroup$
    So if f is one to one it follows that $f^{-1}$ is well defined?
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:33




















  • $begingroup$
    It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
    $endgroup$
    – AnyAD
    Nov 29 '18 at 8:31










  • $begingroup$
    So if f is one to one it follows that $f^{-1}$ is well defined?
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:33


















$begingroup$
It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
$endgroup$
– AnyAD
Nov 29 '18 at 8:31




$begingroup$
It is well defined since $h $ and $f^{-1}$ are. The equality in (2) follows from its definition.
$endgroup$
– AnyAD
Nov 29 '18 at 8:31












$begingroup$
So if f is one to one it follows that $f^{-1}$ is well defined?
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:33






$begingroup$
So if f is one to one it follows that $f^{-1}$ is well defined?
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:33












1 Answer
1






active

oldest

votes


















2












$begingroup$


  1. It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $fbigl([a,b]bigr)bigl(=hbigl([a,b]bigr)bigr)$ into $mathbb R$ and because $hbigl([a,b]bigr)subset D_{f^{-1}}$.

  2. Because $Phi= f^{-1}circ himplies fcircPhi=fcirc(f^{-1}circ h)=h$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you clarify the notation in 1. I don't get all the parenthesis.
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:35










  • $begingroup$
    I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 8:37










  • $begingroup$
    Okey, I am going to have to think about it a bit more before I accept.
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:41






  • 1




    $begingroup$
    But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 8:59






  • 1




    $begingroup$
    An inverse function is always one-to-one.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 9:25











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$


  1. It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $fbigl([a,b]bigr)bigl(=hbigl([a,b]bigr)bigr)$ into $mathbb R$ and because $hbigl([a,b]bigr)subset D_{f^{-1}}$.

  2. Because $Phi= f^{-1}circ himplies fcircPhi=fcirc(f^{-1}circ h)=h$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you clarify the notation in 1. I don't get all the parenthesis.
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:35










  • $begingroup$
    I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 8:37










  • $begingroup$
    Okey, I am going to have to think about it a bit more before I accept.
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:41






  • 1




    $begingroup$
    But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 8:59






  • 1




    $begingroup$
    An inverse function is always one-to-one.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 9:25
















2












$begingroup$


  1. It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $fbigl([a,b]bigr)bigl(=hbigl([a,b]bigr)bigr)$ into $mathbb R$ and because $hbigl([a,b]bigr)subset D_{f^{-1}}$.

  2. Because $Phi= f^{-1}circ himplies fcircPhi=fcirc(f^{-1}circ h)=h$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Could you clarify the notation in 1. I don't get all the parenthesis.
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:35










  • $begingroup$
    I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 8:37










  • $begingroup$
    Okey, I am going to have to think about it a bit more before I accept.
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:41






  • 1




    $begingroup$
    But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 8:59






  • 1




    $begingroup$
    An inverse function is always one-to-one.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 9:25














2












2








2





$begingroup$


  1. It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $fbigl([a,b]bigr)bigl(=hbigl([a,b]bigr)bigr)$ into $mathbb R$ and because $hbigl([a,b]bigr)subset D_{f^{-1}}$.

  2. Because $Phi= f^{-1}circ himplies fcircPhi=fcirc(f^{-1}circ h)=h$.






share|cite|improve this answer









$endgroup$




  1. It is well defined because $f$ is one-to-one and therefore $f^{-1}$ exists as a map from $fbigl([a,b]bigr)bigl(=hbigl([a,b]bigr)bigr)$ into $mathbb R$ and because $hbigl([a,b]bigr)subset D_{f^{-1}}$.

  2. Because $Phi= f^{-1}circ himplies fcircPhi=fcirc(f^{-1}circ h)=h$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 '18 at 8:33









José Carlos SantosJosé Carlos Santos

159k22126231




159k22126231












  • $begingroup$
    Could you clarify the notation in 1. I don't get all the parenthesis.
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:35










  • $begingroup$
    I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 8:37










  • $begingroup$
    Okey, I am going to have to think about it a bit more before I accept.
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:41






  • 1




    $begingroup$
    But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 8:59






  • 1




    $begingroup$
    An inverse function is always one-to-one.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 9:25


















  • $begingroup$
    Could you clarify the notation in 1. I don't get all the parenthesis.
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:35










  • $begingroup$
    I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 8:37










  • $begingroup$
    Okey, I am going to have to think about it a bit more before I accept.
    $endgroup$
    – ALEXANDER
    Nov 29 '18 at 8:41






  • 1




    $begingroup$
    But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 8:59






  • 1




    $begingroup$
    An inverse function is always one-to-one.
    $endgroup$
    – José Carlos Santos
    Nov 29 '18 at 9:25
















$begingroup$
Could you clarify the notation in 1. I don't get all the parenthesis.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:35




$begingroup$
Could you clarify the notation in 1. I don't get all the parenthesis.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:35












$begingroup$
I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:37




$begingroup$
I'm just saying that $fbigl([a,b]bigr)=hbigl([a,b]bigr)$ and that $h$ maps $[a,b]$ into a subset of the domain of $f^{-1}$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:37












$begingroup$
Okey, I am going to have to think about it a bit more before I accept.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:41




$begingroup$
Okey, I am going to have to think about it a bit more before I accept.
$endgroup$
– ALEXANDER
Nov 29 '18 at 8:41




1




1




$begingroup$
But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:59




$begingroup$
But $h(p)$ is equal to $fbigl(Phi(p)bigr)$, for each $pin[a,b]$.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 8:59




1




1




$begingroup$
An inverse function is always one-to-one.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 9:25




$begingroup$
An inverse function is always one-to-one.
$endgroup$
– José Carlos Santos
Nov 29 '18 at 9:25


















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