Example of $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$
$begingroup$
In the pages 135-136 of Algebraic Topology, Hatcher studies the degree of a map $f:S^nto S^n$ in terms of the local degree at the preimages of a point $yin S^n$ such that $f^{-1}(y)$ is a finite set. I wonder if there is for some $n>0$ an example of a map $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$.
Every map I've thought so far has points with finite preimage. In additiom it can be shown that this map can't be differentiable. Therefore it must be a weird map. Does any one know one such example o at least what other properties should it verify?
algebraic-topology continuity
edited Dec 25 '18 at 7:24
Saad
21k92453
asked Dec 24 '18 at 20:53
JaviJavi
3,29321133
$endgroup$
add a comment |
$begingroup$
In the pages 135-136 of Algebraic Topology, Hatcher studies the degree of a map $f:S^nto S^n$ in terms of the local degree at the preimages of a point $yin S^n$ such that $f^{-1}(y)$ is a finite set. I wonder if there is for some $n>0$ an example of a map $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$.
Every map I've thought so far has points with finite preimage. In additiom it can be shown that this map can't be differentiable. Therefore it must be a weird map. Does any one know one such example o at least what other properties should it verify?
algebraic-topology continuity
edited Dec 25 '18 at 7:24
Saad
21k92453
asked Dec 24 '18 at 20:53
JaviJavi
3,29321133
$endgroup$
1
$begingroup$
Think first about maps of intervals constructed via Peano curves.
$endgroup$
– Moishe Kohan
Dec 24 '18 at 21:02
add a comment |
$begingroup$
In the pages 135-136 of Algebraic Topology, Hatcher studies the degree of a map $f:S^nto S^n$ in terms of the local degree at the preimages of a point $yin S^n$ such that $f^{-1}(y)$ is a finite set. I wonder if there is for some $n>0$ an example of a map $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$.
Every map I've thought so far has points with finite preimage. In additiom it can be shown that this map can't be differentiable. Therefore it must be a weird map. Does any one know one such example o at least what other properties should it verify?
algebraic-topology continuity
edited Dec 25 '18 at 7:24
Saad
21k92453
asked Dec 24 '18 at 20:53
JaviJavi
3,29321133
$endgroup$
In the pages 135-136 of Algebraic Topology, Hatcher studies the degree of a map $f:S^nto S^n$ in terms of the local degree at the preimages of a point $yin S^n$ such that $f^{-1}(y)$ is a finite set. I wonder if there is for some $n>0$ an example of a map $f:S^nto S^n$ such that $f^{-1}(y)$ is infinite for all $y$.
Every map I've thought so far has points with finite preimage. In additiom it can be shown that this map can't be differentiable. Therefore it must be a weird map. Does any one know one such example o at least what other properties should it verify?
algebraic-topology continuity
algebraic-topology continuity
edited Dec 25 '18 at 7:24
Saad
21k92453
asked Dec 24 '18 at 20:53
JaviJavi
3,29321133
edited Dec 25 '18 at 7:24
Saad
21k92453
asked Dec 24 '18 at 20:53
JaviJavi
3,29321133
edited Dec 25 '18 at 7:24
Saad
21k92453
edited Dec 25 '18 at 7:24
Saad
21k92453
edited Dec 25 '18 at 7:24
Saad
21k92453
21k92453
asked Dec 24 '18 at 20:53
JaviJavi
3,29321133
asked Dec 24 '18 at 20:53
JaviJavi
3,29321133
asked Dec 24 '18 at 20:53
JaviJavi
3,29321133
3,29321133
1
$begingroup$
Think first about maps of intervals constructed via Peano curves.
$endgroup$
– Moishe Kohan
Dec 24 '18 at 21:02
add a comment |
1
$begingroup$
Think first about maps of intervals constructed via Peano curves.
$endgroup$
– Moishe Kohan
Dec 24 '18 at 21:02
1
1
$begingroup$
Think first about maps of intervals constructed via Peano curves.
$endgroup$
– Moishe Kohan
Dec 24 '18 at 21:02
$begingroup$
Think first about maps of intervals constructed via Peano curves.
$endgroup$
– Moishe Kohan
Dec 24 '18 at 21:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's an example for $n=1$.
Consider a space-filling map $f:[0,1]to[0,1]times[0,1]$, that
is a continuous surjection. Consider $f_1$, the first component of
it. This is a continuous map $[0,1]to[0,1]$ whose fibres are all uncountable.
If we compose with a surjection $[0,1]$ to $S^1$ we get a continuous $g:
[0,1]to S^1$ with all fibres uncountable. Now define
$h:S^1to S^1$ by $h(x,y)=g(|x|)$; then $h$ is a continuous surjection
with all fibres uncountable.
One can use similar tricks for any $n$.
answered Dec 24 '18 at 21:05
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08
add a 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%2f3051626%2fexample-of-fsn-to-sn-such-that-f-1y-is-infinite-for-all-y%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
$begingroup$
Here's an example for $n=1$.
Consider a space-filling map $f:[0,1]to[0,1]times[0,1]$, that
is a continuous surjection. Consider $f_1$, the first component of
it. This is a continuous map $[0,1]to[0,1]$ whose fibres are all uncountable.
If we compose with a surjection $[0,1]$ to $S^1$ we get a continuous $g:
[0,1]to S^1$ with all fibres uncountable. Now define
$h:S^1to S^1$ by $h(x,y)=g(|x|)$; then $h$ is a continuous surjection
with all fibres uncountable.
One can use similar tricks for any $n$.
answered Dec 24 '18 at 21:05
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08
add a comment |
$begingroup$
Here's an example for $n=1$.
Consider a space-filling map $f:[0,1]to[0,1]times[0,1]$, that
is a continuous surjection. Consider $f_1$, the first component of
it. This is a continuous map $[0,1]to[0,1]$ whose fibres are all uncountable.
If we compose with a surjection $[0,1]$ to $S^1$ we get a continuous $g:
[0,1]to S^1$ with all fibres uncountable. Now define
$h:S^1to S^1$ by $h(x,y)=g(|x|)$; then $h$ is a continuous surjection
with all fibres uncountable.
One can use similar tricks for any $n$.
answered Dec 24 '18 at 21:05
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08
add a comment |
$begingroup$
Here's an example for $n=1$.
Consider a space-filling map $f:[0,1]to[0,1]times[0,1]$, that
is a continuous surjection. Consider $f_1$, the first component of
it. This is a continuous map $[0,1]to[0,1]$ whose fibres are all uncountable.
If we compose with a surjection $[0,1]$ to $S^1$ we get a continuous $g:
[0,1]to S^1$ with all fibres uncountable. Now define
$h:S^1to S^1$ by $h(x,y)=g(|x|)$; then $h$ is a continuous surjection
with all fibres uncountable.
One can use similar tricks for any $n$.
answered Dec 24 '18 at 21:05
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
Here's an example for $n=1$.
Consider a space-filling map $f:[0,1]to[0,1]times[0,1]$, that
is a continuous surjection. Consider $f_1$, the first component of
it. This is a continuous map $[0,1]to[0,1]$ whose fibres are all uncountable.
If we compose with a surjection $[0,1]$ to $S^1$ we get a continuous $g:
[0,1]to S^1$ with all fibres uncountable. Now define
$h:S^1to S^1$ by $h(x,y)=g(|x|)$; then $h$ is a continuous surjection
with all fibres uncountable.
One can use similar tricks for any $n$.
answered Dec 24 '18 at 21:05
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Dec 24 '18 at 21:05
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Dec 24 '18 at 21:05
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Dec 24 '18 at 21:05
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08
add a comment |
$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08
$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08
$begingroup$
It wasn't as hard as I thought! Thanks
$endgroup$
– Javi
Dec 24 '18 at 21:08
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.
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%2f3051626%2fexample-of-fsn-to-sn-such-that-f-1y-is-infinite-for-all-y%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
$begingroup$
Think first about maps of intervals constructed via Peano curves.
$endgroup$
– Moishe Kohan
Dec 24 '18 at 21:02