Inverse function being onto.
Follow up on this question:
As you can see from this link
$Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ now in my textbook the next theorem following what I posted in the link, is as follows:
Theorem 2
Let $f:[a,b]subseteq Rto R^n$ and $h:[c,d]subseteq Rto R^n$ be 1-1 parametrizations of a simple arc C in $R^n$ Then there exists a unique function $Phi$ from [a,b] onto [c,d] such that $h=fcircPhi$. Moreover, $Phi$ is continuous and strictly monotonic.
Now I am getting confused as the proof provided for theorem 2 is the teorem provided in the link( 'Follow up question') but in the follow up question $Phi$ maps [c,d] onto [a,b] and not opposite. What am I missing?
to be more specific the proof states:
By Theorem 1(link) the function $Phi$ is 1-1, continuous and strictly monotonic.
proof-explanation multilinear-algebra
asked Nov 29 at 11:56
ALEXANDER
8781921
|
show 4 more comments
Follow up on this question:
As you can see from this link
$Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ now in my textbook the next theorem following what I posted in the link, is as follows:
Theorem 2
Let $f:[a,b]subseteq Rto R^n$ and $h:[c,d]subseteq Rto R^n$ be 1-1 parametrizations of a simple arc C in $R^n$ Then there exists a unique function $Phi$ from [a,b] onto [c,d] such that $h=fcircPhi$. Moreover, $Phi$ is continuous and strictly monotonic.
Now I am getting confused as the proof provided for theorem 2 is the teorem provided in the link( 'Follow up question') but in the follow up question $Phi$ maps [c,d] onto [a,b] and not opposite. What am I missing?
to be more specific the proof states:
By Theorem 1(link) the function $Phi$ is 1-1, continuous and strictly monotonic.
proof-explanation multilinear-algebra
asked Nov 29 at 11:56
ALEXANDER
8781921
2
$phi$ should map $[c,d]$ onto $[a,b]. h$ is defined on $[c,d]$, and $f$ is defined over $[a,b]$, so $phi: [c,d]rightarrow[a,b], phi(t)=frac{a-b}{c-d}t+frac{bc-ad}{c-d}$
– Shubham Johri
Nov 29 at 12:10
1
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 at 12:13
1
Note that it is generally not true that a curve will span the entire codomain $mathbb R^n$. When we talk about the invertibility of $f$, we are talking about $f$ being surjective on its range, $f([a,b])$, which is always true. Also, since $phi$ maps $[c,d]$ onto $[a,b]$, the range of $f$ and $h=fcircphi$ is identical. So, the case of $f$ and $h$ having different ranges doesn't arise.
– Shubham Johri
Nov 29 at 12:24
1
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 at 12:37
1
That is one way of seeing it. You could also say that every point in $[a,b]$ is the image of some point in $[c,d]$ under $phi$ i.e. $R(phi)=phi([c,d])=[a,b]$, so the range of $h=h([c,d])=f(phi([c,d]))=f([a,b])=R(f)$, the range of $f$.
– Shubham Johri
Nov 29 at 13:32
|
show 4 more comments
Follow up on this question:
As you can see from this link
$Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ now in my textbook the next theorem following what I posted in the link, is as follows:
Theorem 2
Let $f:[a,b]subseteq Rto R^n$ and $h:[c,d]subseteq Rto R^n$ be 1-1 parametrizations of a simple arc C in $R^n$ Then there exists a unique function $Phi$ from [a,b] onto [c,d] such that $h=fcircPhi$. Moreover, $Phi$ is continuous and strictly monotonic.
Now I am getting confused as the proof provided for theorem 2 is the teorem provided in the link( 'Follow up question') but in the follow up question $Phi$ maps [c,d] onto [a,b] and not opposite. What am I missing?
to be more specific the proof states:
By Theorem 1(link) the function $Phi$ is 1-1, continuous and strictly monotonic.
proof-explanation multilinear-algebra
asked Nov 29 at 11:56
ALEXANDER
8781921
Follow up on this question:
As you can see from this link
$Phi=f^{-1}circ h$ is well defined and maps $[c,d]$ onto $[a,b]$ now in my textbook the next theorem following what I posted in the link, is as follows:
Theorem 2
Let $f:[a,b]subseteq Rto R^n$ and $h:[c,d]subseteq Rto R^n$ be 1-1 parametrizations of a simple arc C in $R^n$ Then there exists a unique function $Phi$ from [a,b] onto [c,d] such that $h=fcircPhi$. Moreover, $Phi$ is continuous and strictly monotonic.
Now I am getting confused as the proof provided for theorem 2 is the teorem provided in the link( 'Follow up question') but in the follow up question $Phi$ maps [c,d] onto [a,b] and not opposite. What am I missing?
to be more specific the proof states:
By Theorem 1(link) the function $Phi$ is 1-1, continuous and strictly monotonic.
proof-explanation multilinear-algebra
proof-explanation multilinear-algebra
asked Nov 29 at 11:56
ALEXANDER
8781921
asked Nov 29 at 11:56
ALEXANDER
8781921
asked Nov 29 at 11:56
ALEXANDER
8781921
asked Nov 29 at 11:56
ALEXANDER
8781921
asked Nov 29 at 11:56
ALEXANDER
8781921
8781921
2
$phi$ should map $[c,d]$ onto $[a,b]. h$ is defined on $[c,d]$, and $f$ is defined over $[a,b]$, so $phi: [c,d]rightarrow[a,b], phi(t)=frac{a-b}{c-d}t+frac{bc-ad}{c-d}$
– Shubham Johri
Nov 29 at 12:10
1
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 at 12:13
1
Note that it is generally not true that a curve will span the entire codomain $mathbb R^n$. When we talk about the invertibility of $f$, we are talking about $f$ being surjective on its range, $f([a,b])$, which is always true. Also, since $phi$ maps $[c,d]$ onto $[a,b]$, the range of $f$ and $h=fcircphi$ is identical. So, the case of $f$ and $h$ having different ranges doesn't arise.
– Shubham Johri
Nov 29 at 12:24
1
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 at 12:37
1
That is one way of seeing it. You could also say that every point in $[a,b]$ is the image of some point in $[c,d]$ under $phi$ i.e. $R(phi)=phi([c,d])=[a,b]$, so the range of $h=h([c,d])=f(phi([c,d]))=f([a,b])=R(f)$, the range of $f$.
– Shubham Johri
Nov 29 at 13:32
|
show 4 more comments
2
$phi$ should map $[c,d]$ onto $[a,b]. h$ is defined on $[c,d]$, and $f$ is defined over $[a,b]$, so $phi: [c,d]rightarrow[a,b], phi(t)=frac{a-b}{c-d}t+frac{bc-ad}{c-d}$
– Shubham Johri
Nov 29 at 12:10
1
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 at 12:13
1
Note that it is generally not true that a curve will span the entire codomain $mathbb R^n$. When we talk about the invertibility of $f$, we are talking about $f$ being surjective on its range, $f([a,b])$, which is always true. Also, since $phi$ maps $[c,d]$ onto $[a,b]$, the range of $f$ and $h=fcircphi$ is identical. So, the case of $f$ and $h$ having different ranges doesn't arise.
– Shubham Johri
Nov 29 at 12:24
1
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 at 12:37
1
That is one way of seeing it. You could also say that every point in $[a,b]$ is the image of some point in $[c,d]$ under $phi$ i.e. $R(phi)=phi([c,d])=[a,b]$, so the range of $h=h([c,d])=f(phi([c,d]))=f([a,b])=R(f)$, the range of $f$.
– Shubham Johri
Nov 29 at 13:32
2
2
$phi$ should map $[c,d]$ onto $[a,b]. h$ is defined on $[c,d]$, and $f$ is defined over $[a,b]$, so $phi: [c,d]rightarrow[a,b], phi(t)=frac{a-b}{c-d}t+frac{bc-ad}{c-d}$
– Shubham Johri
Nov 29 at 12:10
$phi$ should map $[c,d]$ onto $[a,b]. h$ is defined on $[c,d]$, and $f$ is defined over $[a,b]$, so $phi: [c,d]rightarrow[a,b], phi(t)=frac{a-b}{c-d}t+frac{bc-ad}{c-d}$
– Shubham Johri
Nov 29 at 12:10
1
1
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 at 12:13
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 at 12:13
1
1
Note that it is generally not true that a curve will span the entire codomain $mathbb R^n$. When we talk about the invertibility of $f$, we are talking about $f$ being surjective on its range, $f([a,b])$, which is always true. Also, since $phi$ maps $[c,d]$ onto $[a,b]$, the range of $f$ and $h=fcircphi$ is identical. So, the case of $f$ and $h$ having different ranges doesn't arise.
– Shubham Johri
Nov 29 at 12:24
Note that it is generally not true that a curve will span the entire codomain $mathbb R^n$. When we talk about the invertibility of $f$, we are talking about $f$ being surjective on its range, $f([a,b])$, which is always true. Also, since $phi$ maps $[c,d]$ onto $[a,b]$, the range of $f$ and $h=fcircphi$ is identical. So, the case of $f$ and $h$ having different ranges doesn't arise.
– Shubham Johri
Nov 29 at 12:24
1
1
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 at 12:37
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 at 12:37
1
1
That is one way of seeing it. You could also say that every point in $[a,b]$ is the image of some point in $[c,d]$ under $phi$ i.e. $R(phi)=phi([c,d])=[a,b]$, so the range of $h=h([c,d])=f(phi([c,d]))=f([a,b])=R(f)$, the range of $f$.
– Shubham Johri
Nov 29 at 13:32
That is one way of seeing it. You could also say that every point in $[a,b]$ is the image of some point in $[c,d]$ under $phi$ i.e. $R(phi)=phi([c,d])=[a,b]$, so the range of $h=h([c,d])=f(phi([c,d]))=f([a,b])=R(f)$, the range of $f$.
– Shubham Johri
Nov 29 at 13:32
|
show 4 more comments
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2
$phi$ should map $[c,d]$ onto $[a,b]. h$ is defined on $[c,d]$, and $f$ is defined over $[a,b]$, so $phi: [c,d]rightarrow[a,b], phi(t)=frac{a-b}{c-d}t+frac{bc-ad}{c-d}$
– Shubham Johri
Nov 29 at 12:10
1
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 at 12:13
1
Note that it is generally not true that a curve will span the entire codomain $mathbb R^n$. When we talk about the invertibility of $f$, we are talking about $f$ being surjective on its range, $f([a,b])$, which is always true. Also, since $phi$ maps $[c,d]$ onto $[a,b]$, the range of $f$ and $h=fcircphi$ is identical. So, the case of $f$ and $h$ having different ranges doesn't arise.
– Shubham Johri
Nov 29 at 12:24
1
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 at 12:37
1
That is one way of seeing it. You could also say that every point in $[a,b]$ is the image of some point in $[c,d]$ under $phi$ i.e. $R(phi)=phi([c,d])=[a,b]$, so the range of $h=h([c,d])=f(phi([c,d]))=f([a,b])=R(f)$, the range of $f$.
– Shubham Johri
Nov 29 at 13:32