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 '18 at 11:56
ALEXANDER
8881921
|
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 '18 at 11:56
ALEXANDER
8881921
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 '18 at 12:10
1
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 '18 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 '18 at 12:24
1
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 '18 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 '18 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 '18 at 11:56
ALEXANDER
8881921
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 '18 at 11:56
ALEXANDER
8881921
asked Nov 29 '18 at 11:56
ALEXANDER
8881921
asked Nov 29 '18 at 11:56
ALEXANDER
8881921
asked Nov 29 '18 at 11:56
ALEXANDER
8881921
asked Nov 29 '18 at 11:56
ALEXANDER
8881921
8881921
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 '18 at 12:10
1
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 '18 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 '18 at 12:24
1
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 '18 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 '18 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 '18 at 12:10
1
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 '18 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 '18 at 12:24
1
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 '18 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 '18 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 '18 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 '18 at 12:10
1
1
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 '18 at 12:13
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 '18 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 '18 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 '18 at 12:24
1
1
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 '18 at 12:37
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 '18 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 '18 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 '18 at 13:32
|
show 4 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018528%2finverse-function-being-onto%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018528%2finverse-function-being-onto%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 '18 at 12:10
1
I suppose that you have spotted an error on your textbook.
– José Carlos Santos
Nov 29 '18 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 '18 at 12:24
1
That is true. Their range is the curve $mathbb C$.
– Shubham Johri
Nov 29 '18 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 '18 at 13:32