Do i imagine the linear (straight line) homotopy in a correct way? Announcing the arrival of...
My mentor says to set image to Fine instead of RAW — how is this different from JPG?
What is the proper term for etching or digging of wall to hide conduit of cables
In musical terms, what properties are varied by the human voice to produce different words / syllables?
2018 MacBook Pro won't let me install macOS High Sierra 10.13 from USB installer
Is there a spell that can create a permanent fire?
Did any compiler fully use 80-bit floating point?
Does a random sequence of vectors span a Hilbert space?
What is a more techy Technical Writer job title that isn't cutesy or confusing?
Why are two-digit numbers in Jonathan Swift's "Gulliver's Travels" (1726) written in "German style"?
Is the Mordenkainens' Sword spell underpowered?
Does the main washing effect of soap come from foam?
Adapting the Chinese Remainder Theorem (CRT) for integers to polynomials
Keep at all times, the minus sign above aligned with minus sign below
Does the universe have a fixed centre of mass?
Twin's vs. Twins'
One-one communication
How many time has Arya actually used Needle?
Pointing to problems without suggesting solutions
How can I list files in reverse time order by a command and pass them as arguments to another command?
Was the pager message from Nick Fury to Captain Marvel unnecessary?
Why does BitLocker not use RSA?
Why is there so little support for joining EFTA in the British parliament?
Where and when has Thucydides been studied?
Marquee sign letters
Do i imagine the linear (straight line) homotopy in a correct way?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)simple proposition about homotopy groupsymmetry in the homotopy relationDetermining images of points in a path homotopy.Difference between two concepts of homotopy for simplicial maps?Maps to Sn homotopicShow that g*c and c*g are homotopic, where g is a loop and c a constant loopComputing fundamental group of the complement to three infinite straight lines, and of complement to $S^1 cup {Z} $Linking numbers.tensor product and loop space multiplication are homotopic on $mathbb{P}^infty$Star convex subset $A$ of $mathbb{R}^n$ is simply connected
$begingroup$
Today i learned about the linear homotopy which says that any two paths $f_0, f_1$ in $mathbb{R}^n$ are homotopic via the homotopy $$ f_t(s) = (1-t)f_0(s) + tf_1(s)$$
Am i right in imagining the given homotopy as something like this?
such that $F(s,t) = f_t(s) $ are simply the linesegments going from $f_0(s)$ towards $f_1(s)$ as the straight lines (which "connect" $f_0$ and $f_1$ for every $sin [0,1]$) as drawn in the picture?
Sorry if this question might be a trivial one, i just want to make sure i don't get things wrong.
Thanks for any kind of feedback!
algebraic-topology homotopy-theory path-connected
asked 1 hour ago
ZestZest
301213
$endgroup$
add a comment |
$begingroup$
Today i learned about the linear homotopy which says that any two paths $f_0, f_1$ in $mathbb{R}^n$ are homotopic via the homotopy $$ f_t(s) = (1-t)f_0(s) + tf_1(s)$$
Am i right in imagining the given homotopy as something like this?
such that $F(s,t) = f_t(s) $ are simply the linesegments going from $f_0(s)$ towards $f_1(s)$ as the straight lines (which "connect" $f_0$ and $f_1$ for every $sin [0,1]$) as drawn in the picture?
Sorry if this question might be a trivial one, i just want to make sure i don't get things wrong.
Thanks for any kind of feedback!
algebraic-topology homotopy-theory path-connected
asked 1 hour ago
ZestZest
301213
$endgroup$
1
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
1
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago
add a comment |
$begingroup$
Today i learned about the linear homotopy which says that any two paths $f_0, f_1$ in $mathbb{R}^n$ are homotopic via the homotopy $$ f_t(s) = (1-t)f_0(s) + tf_1(s)$$
Am i right in imagining the given homotopy as something like this?
such that $F(s,t) = f_t(s) $ are simply the linesegments going from $f_0(s)$ towards $f_1(s)$ as the straight lines (which "connect" $f_0$ and $f_1$ for every $sin [0,1]$) as drawn in the picture?
Sorry if this question might be a trivial one, i just want to make sure i don't get things wrong.
Thanks for any kind of feedback!
algebraic-topology homotopy-theory path-connected
asked 1 hour ago
ZestZest
301213
$endgroup$
Today i learned about the linear homotopy which says that any two paths $f_0, f_1$ in $mathbb{R}^n$ are homotopic via the homotopy $$ f_t(s) = (1-t)f_0(s) + tf_1(s)$$
Am i right in imagining the given homotopy as something like this?
such that $F(s,t) = f_t(s) $ are simply the linesegments going from $f_0(s)$ towards $f_1(s)$ as the straight lines (which "connect" $f_0$ and $f_1$ for every $sin [0,1]$) as drawn in the picture?
Sorry if this question might be a trivial one, i just want to make sure i don't get things wrong.
Thanks for any kind of feedback!
algebraic-topology homotopy-theory path-connected
algebraic-topology homotopy-theory path-connected
asked 1 hour ago
ZestZest
301213
asked 1 hour ago
ZestZest
301213
asked 1 hour ago
ZestZest
301213
asked 1 hour ago
ZestZest
301213
asked 1 hour ago
ZestZest
301213
301213
1
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
1
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago
add a comment |
1
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
1
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago
1
1
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
1
1
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $bar{s}$ and call $a := f_0(bar{s})$ and $b := f_1(bar{s})$. Examining the homotopy,
$$
H_{bar{s}}(t) := F(bar{s}, t) = a(1-t) + bt
$$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $mathbb{R}^n$. If you draw this line for each $bar{s}$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant,
$$
dfrac{dH_{bar{s}}}{dt} = b - a
$$
answered 1 hour ago
jnez71jnez71
2,495720
$endgroup$
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
add a comment |
$begingroup$
Yes, it is absolutely correct.
$endgroup$
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
|
show 1 more comment
Your Answer
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%2f3196440%2fdo-i-imagine-the-linear-straight-line-homotopy-in-a-correct-way%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $bar{s}$ and call $a := f_0(bar{s})$ and $b := f_1(bar{s})$. Examining the homotopy,
$$
H_{bar{s}}(t) := F(bar{s}, t) = a(1-t) + bt
$$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $mathbb{R}^n$. If you draw this line for each $bar{s}$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant,
$$
dfrac{dH_{bar{s}}}{dt} = b - a
$$
answered 1 hour ago
jnez71jnez71
2,495720
$endgroup$
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
add a comment |
$begingroup$
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $bar{s}$ and call $a := f_0(bar{s})$ and $b := f_1(bar{s})$. Examining the homotopy,
$$
H_{bar{s}}(t) := F(bar{s}, t) = a(1-t) + bt
$$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $mathbb{R}^n$. If you draw this line for each $bar{s}$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant,
$$
dfrac{dH_{bar{s}}}{dt} = b - a
$$
answered 1 hour ago
jnez71jnez71
2,495720
$endgroup$
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
add a comment |
$begingroup$
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $bar{s}$ and call $a := f_0(bar{s})$ and $b := f_1(bar{s})$. Examining the homotopy,
$$
H_{bar{s}}(t) := F(bar{s}, t) = a(1-t) + bt
$$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $mathbb{R}^n$. If you draw this line for each $bar{s}$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant,
$$
dfrac{dH_{bar{s}}}{dt} = b - a
$$
answered 1 hour ago
jnez71jnez71
2,495720
$endgroup$
Yes. Perhaps to help you see why this is true, pick an arbitrary $s$ value $bar{s}$ and call $a := f_0(bar{s})$ and $b := f_1(bar{s})$. Examining the homotopy,
$$
H_{bar{s}}(t) := F(bar{s}, t) = a(1-t) + bt
$$
we see that it is the parametric equation of a straight line connecting points $a$ and $b$ in $mathbb{R}^n$. If you draw this line for each $bar{s}$ choice, you get your diagram. Perhaps write a bit of code to construct such a plot?
Furthermore, one may be interested in the "speed" at which $a$ "moves" to $b$. We find it to be a constant,
$$
dfrac{dH_{bar{s}}}{dt} = b - a
$$
answered 1 hour ago
jnez71jnez71
2,495720
edited 1 hour ago
answered 1 hour ago
jnez71jnez71
2,495720
answered 1 hour ago
jnez71jnez71
2,495720
answered 1 hour ago
jnez71jnez71
2,495720
2,495720
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
add a comment |
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
$begingroup$
this is very helpful @jnez71. Thanks a lot!
$endgroup$
– Zest
1 hour ago
1
1
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
$begingroup$
No problem! I added a bit more to cover some of the comments about speed
$endgroup$
– jnez71
1 hour ago
add a comment |
$begingroup$
Yes, it is absolutely correct.
$endgroup$
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
|
show 1 more comment
$begingroup$
Yes, it is absolutely correct.
$endgroup$
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
|
show 1 more comment
$begingroup$
Yes, it is absolutely correct.
$endgroup$
Yes, it is absolutely correct.
answered 1 hour ago
community wiki
Paul Frost
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
|
show 1 more comment
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
thank you very much @Paul. Just being curious, what does it mean that your answer is "community wiki"?
$endgroup$
– Zest
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
See math.stackexchange.com/help/privileges/edit-community-wiki. A community wiki can be edited by everybody. Frankly, my answer was only a feedback, and I guess you didn't expect more. But of course it doesn't deserve positive reputation.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
$begingroup$
thanks Paul. In fact, your answer was all i was hoping for. Would you mind me accepting @jnez72's answer rather than yours even though both helped me?
$endgroup$
– Zest
1 hour ago
1
1
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
No problem - it is okay!
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
$begingroup$
By the way, in my opinion it is important to make it visible in the question queue that a question has been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "248,877 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. If that should happen to one of your questions, do not hesitate to write an answer and acknowledge it.
$endgroup$
– Paul Frost
1 hour ago
|
show 1 more comment
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.
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%2f3196440%2fdo-i-imagine-the-linear-straight-line-homotopy-in-a-correct-way%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
1
$begingroup$
It's correct. Furthermore, any point $s$ moves along its straight line at constant speed.
$endgroup$
– Lukas Kofler
1 hour ago
$begingroup$
isn't the point $t$ rather moving along its straight line? to me, $s$ is the parameter moving along the lines from $x_0$ to $x_1$ whereas $t$ moves along the straight linesegments between $f_0$ and $f_1$
$endgroup$
– Zest
1 hour ago
1
$begingroup$
That's what I was trying to say, sorry -- a fixed point $f_0(s)$ moves at constant speed as $t$ varies.
$endgroup$
– Lukas Kofler
1 hour ago