All quasi-isometry is coarsely surjective.
Definition. $(X,d_X)$, $(Y,d_Y)$ metric spaces.
$f:Xto Y$ is quasi-isometry if
$d_Y(f(x),f(x'))leq Ld_X(x,x')+C$
exists coarse inverse, i.e. $overline{f}:Yto X$ such that
$d_X(overline{f}f(x),x)leq C$
$d_Y(foverline{f}(y),y)leq C$
and $d_X(overline{f}(y),overline{f}(y'))leq Ld_Y(y,y')+C$
for all $x,x'in X$, for all $y,y'in Y$
How prove that $f:Xto Y$ ($L-C$) quasi-isometry then $f$ is coarsely surjective? i.e. $forall yin Y, exists f(x)in f(X)$ such that $d_Y(y,f(x))leq C$
metric-spaces lipschitz-functions
asked Nov 7 '18 at 15:53
eraldcoil
370111
add a comment |
Definition. $(X,d_X)$, $(Y,d_Y)$ metric spaces.
$f:Xto Y$ is quasi-isometry if
$d_Y(f(x),f(x'))leq Ld_X(x,x')+C$
exists coarse inverse, i.e. $overline{f}:Yto X$ such that
$d_X(overline{f}f(x),x)leq C$
$d_Y(foverline{f}(y),y)leq C$
and $d_X(overline{f}(y),overline{f}(y'))leq Ld_Y(y,y')+C$
for all $x,x'in X$, for all $y,y'in Y$
How prove that $f:Xto Y$ ($L-C$) quasi-isometry then $f$ is coarsely surjective? i.e. $forall yin Y, exists f(x)in f(X)$ such that $d_Y(y,f(x))leq C$
metric-spaces lipschitz-functions
asked Nov 7 '18 at 15:53
eraldcoil
370111
add a comment |
Definition. $(X,d_X)$, $(Y,d_Y)$ metric spaces.
$f:Xto Y$ is quasi-isometry if
$d_Y(f(x),f(x'))leq Ld_X(x,x')+C$
exists coarse inverse, i.e. $overline{f}:Yto X$ such that
$d_X(overline{f}f(x),x)leq C$
$d_Y(foverline{f}(y),y)leq C$
and $d_X(overline{f}(y),overline{f}(y'))leq Ld_Y(y,y')+C$
for all $x,x'in X$, for all $y,y'in Y$
How prove that $f:Xto Y$ ($L-C$) quasi-isometry then $f$ is coarsely surjective? i.e. $forall yin Y, exists f(x)in f(X)$ such that $d_Y(y,f(x))leq C$
metric-spaces lipschitz-functions
asked Nov 7 '18 at 15:53
eraldcoil
370111
Definition. $(X,d_X)$, $(Y,d_Y)$ metric spaces.
$f:Xto Y$ is quasi-isometry if
$d_Y(f(x),f(x'))leq Ld_X(x,x')+C$
exists coarse inverse, i.e. $overline{f}:Yto X$ such that
$d_X(overline{f}f(x),x)leq C$
$d_Y(foverline{f}(y),y)leq C$
and $d_X(overline{f}(y),overline{f}(y'))leq Ld_Y(y,y')+C$
for all $x,x'in X$, for all $y,y'in Y$
How prove that $f:Xto Y$ ($L-C$) quasi-isometry then $f$ is coarsely surjective? i.e. $forall yin Y, exists f(x)in f(X)$ such that $d_Y(y,f(x))leq C$
metric-spaces lipschitz-functions
metric-spaces lipschitz-functions
asked Nov 7 '18 at 15:53
eraldcoil
370111
asked Nov 7 '18 at 15:53
eraldcoil
370111
edited Nov 7 '18 at 15:58
asked Nov 7 '18 at 15:53
eraldcoil
370111
asked Nov 7 '18 at 15:53
eraldcoil
370111
asked Nov 7 '18 at 15:53
eraldcoil
370111
370111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Put $x=bar f(y)$.
..........
Given $yin Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=bar f(y)$ fits.
answered Dec 5 '18 at 8:11
Alex Ravsky
39.3k32181
add a comment |
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%2f2988703%2fall-quasi-isometry-is-coarsely-surjective%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Put $x=bar f(y)$.
..........
Given $yin Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=bar f(y)$ fits.
answered Dec 5 '18 at 8:11
Alex Ravsky
39.3k32181
add a comment |
Put $x=bar f(y)$.
..........
Given $yin Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=bar f(y)$ fits.
answered Dec 5 '18 at 8:11
Alex Ravsky
39.3k32181
add a comment |
Put $x=bar f(y)$.
..........
Given $yin Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=bar f(y)$ fits.
answered Dec 5 '18 at 8:11
Alex Ravsky
39.3k32181
Put $x=bar f(y)$.
..........
Given $yin Y$ we need to find $x$ satisfying given property. But the definition of a quasi-isometry trivially implies that $x=bar f(y)$ fits.
answered Dec 5 '18 at 8:11
Alex Ravsky
39.3k32181
edited 2 days ago
answered Dec 5 '18 at 8:11
Alex Ravsky
39.3k32181
answered Dec 5 '18 at 8:11
Alex Ravsky
39.3k32181
answered Dec 5 '18 at 8:11
Alex Ravsky
39.3k32181
39.3k32181
add a comment |
add a 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.
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%2f2988703%2fall-quasi-isometry-is-coarsely-surjective%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
