Mixing
Path Connectedness and continuous bijections
$begingroup$
Mathoverflow.
Are there any two topological spaces $X$ and $Y$ such that they are path connected and such that there exist continuous bijections $Xrightarrow Y$ and $Yrightarrow X$, but and yet they are not homeomorphic?
Without Path-Connectedness requirement, this is easily fulfilled as the examples in the cited post.
If it indeed implies homeomorphism, how can I prove it?
general-topology continuity connectedness abstract-algebra
edited Apr 13 '17 at 12:58
Community♦
1
$endgroup$
add a comment |
$begingroup$
Mathoverflow.
Are there any two topological spaces $X$ and $Y$ such that they are path connected and such that there exist continuous bijections $Xrightarrow Y$ and $Yrightarrow X$, but and yet they are not homeomorphic?
Without Path-Connectedness requirement, this is easily fulfilled as the examples in the cited post.
If it indeed implies homeomorphism, how can I prove it?
general-topology continuity connectedness abstract-algebra
edited Apr 13 '17 at 12:58
Community♦
1
$endgroup$
1
$begingroup$
My mistake: I overlooked that you wanted a map in both directions. Please ignore my vote for closure.
$endgroup$
– Martin
Apr 9 '13 at 14:28
add a comment |
$begingroup$
Mathoverflow.
Are there any two topological spaces $X$ and $Y$ such that they are path connected and such that there exist continuous bijections $Xrightarrow Y$ and $Yrightarrow X$, but and yet they are not homeomorphic?
Without Path-Connectedness requirement, this is easily fulfilled as the examples in the cited post.
If it indeed implies homeomorphism, how can I prove it?
general-topology continuity connectedness abstract-algebra
edited Apr 13 '17 at 12:58
Community♦
1
$endgroup$
Mathoverflow.
Are there any two topological spaces $X$ and $Y$ such that they are path connected and such that there exist continuous bijections $Xrightarrow Y$ and $Yrightarrow X$, but and yet they are not homeomorphic?
Without Path-Connectedness requirement, this is easily fulfilled as the examples in the cited post.
If it indeed implies homeomorphism, how can I prove it?
general-topology continuity connectedness abstract-algebra
general-topology continuity connectedness abstract-algebra
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
1
asked Apr 9 '13 at 8:58
user45099
1
$begingroup$
My mistake: I overlooked that you wanted a map in both directions. Please ignore my vote for closure.
$endgroup$
– Martin
Apr 9 '13 at 14:28
add a comment |
1
$begingroup$
My mistake: I overlooked that you wanted a map in both directions. Please ignore my vote for closure.
$endgroup$
– Martin
Apr 9 '13 at 14:28
1
1
$begingroup$
My mistake: I overlooked that you wanted a map in both directions. Please ignore my vote for closure.
$endgroup$
– Martin
Apr 9 '13 at 14:28
$begingroup$
My mistake: I overlooked that you wanted a map in both directions. Please ignore my vote for closure.
$endgroup$
– Martin
Apr 9 '13 at 14:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
See the example in my question here. There maps described there are continuous bijections and they induce continuous bijections between the cones of $X$ and $Y$. And cones are path-connected since any point can be joined to the apex.
$endgroup$
1
$begingroup$
Haha, I was wondering why this very similar question to yours appeared at the top of the page.
$endgroup$
– Dan Rust
Dec 13 '13 at 13:39
add a comment |
$begingroup$
Yes, there are. But first lets consider criteria for homeomorphisms.
let $psi$$:Xrightarrow Y$ be a continuous bijection where $X$ is compact and $Y$ is Hausdorff. This is a homeomorphism. This means that topological invariants such as path-connectedness are preserved. So if we DO have a continuous bijection, and $X$ is compact and path-connected, and $Y$ is Hausdorff, then $f(X)subseteq Y$ is path-connected.
Now to answer your question directly. Let $X=mathbb{R}^2$ and $Y=mathbb{R}$ with the standard topology. Both of these spaces are path connected. We can construct a bijection between the two. Let $psi(x,y)=||x+y||-||x-y||$
this map is clearly surjective. It is also clearly injective. So it is bijective, and we know the norm function is continuous. So this is a continuous bijection, both of $mathbb{R}^2$ and $mathbb{R}$ are path-connected, but is well known that $X$ and $Y$ are not homeomorphic to each other.
Why don't we have a homemorphism? We have a continuous bijection, $X$ and $Y$ are Hausdorff, but $mathbb{R}^2$ is not compact. So the assumption of compactness is essential.
answered Oct 24 '16 at 2:29
Kernel_DirichletKernel_Dirichlet
1,138416
$endgroup$
1
$begingroup$
Your function is symmetric in both arguments so it is certainly not bijective
$endgroup$
– ThorbenK
Jan 8 at 17:01
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%2f355763%2fpath-connectedness-and-continuous-bijections%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$
See the example in my question here. There maps described there are continuous bijections and they induce continuous bijections between the cones of $X$ and $Y$. And cones are path-connected since any point can be joined to the apex.
$endgroup$
1
$begingroup$
Haha, I was wondering why this very similar question to yours appeared at the top of the page.
$endgroup$
– Dan Rust
Dec 13 '13 at 13:39
add a comment |
$begingroup$
See the example in my question here. There maps described there are continuous bijections and they induce continuous bijections between the cones of $X$ and $Y$. And cones are path-connected since any point can be joined to the apex.
$endgroup$
1
$begingroup$
Haha, I was wondering why this very similar question to yours appeared at the top of the page.
$endgroup$
– Dan Rust
Dec 13 '13 at 13:39
add a comment |
$begingroup$
See the example in my question here. There maps described there are continuous bijections and they induce continuous bijections between the cones of $X$ and $Y$. And cones are path-connected since any point can be joined to the apex.
$endgroup$
See the example in my question here. There maps described there are continuous bijections and they induce continuous bijections between the cones of $X$ and $Y$. And cones are path-connected since any point can be joined to the apex.
edited Apr 13 '17 at 12:20
community wiki
2 revs
Stefan Hamcke
1
$begingroup$
Haha, I was wondering why this very similar question to yours appeared at the top of the page.
$endgroup$
– Dan Rust
Dec 13 '13 at 13:39
add a comment |
1
$begingroup$
Haha, I was wondering why this very similar question to yours appeared at the top of the page.
$endgroup$
– Dan Rust
Dec 13 '13 at 13:39
1
1
$begingroup$
Haha, I was wondering why this very similar question to yours appeared at the top of the page.
$endgroup$
– Dan Rust
Dec 13 '13 at 13:39
$begingroup$
Haha, I was wondering why this very similar question to yours appeared at the top of the page.
$endgroup$
– Dan Rust
Dec 13 '13 at 13:39
add a comment |
$begingroup$
Yes, there are. But first lets consider criteria for homeomorphisms.
let $psi$$:Xrightarrow Y$ be a continuous bijection where $X$ is compact and $Y$ is Hausdorff. This is a homeomorphism. This means that topological invariants such as path-connectedness are preserved. So if we DO have a continuous bijection, and $X$ is compact and path-connected, and $Y$ is Hausdorff, then $f(X)subseteq Y$ is path-connected.
Now to answer your question directly. Let $X=mathbb{R}^2$ and $Y=mathbb{R}$ with the standard topology. Both of these spaces are path connected. We can construct a bijection between the two. Let $psi(x,y)=||x+y||-||x-y||$
this map is clearly surjective. It is also clearly injective. So it is bijective, and we know the norm function is continuous. So this is a continuous bijection, both of $mathbb{R}^2$ and $mathbb{R}$ are path-connected, but is well known that $X$ and $Y$ are not homeomorphic to each other.
Why don't we have a homemorphism? We have a continuous bijection, $X$ and $Y$ are Hausdorff, but $mathbb{R}^2$ is not compact. So the assumption of compactness is essential.
answered Oct 24 '16 at 2:29
Kernel_DirichletKernel_Dirichlet
1,138416
$endgroup$
1
$begingroup$
Your function is symmetric in both arguments so it is certainly not bijective
$endgroup$
– ThorbenK
Jan 8 at 17:01
add a comment |
$begingroup$
Yes, there are. But first lets consider criteria for homeomorphisms.
let $psi$$:Xrightarrow Y$ be a continuous bijection where $X$ is compact and $Y$ is Hausdorff. This is a homeomorphism. This means that topological invariants such as path-connectedness are preserved. So if we DO have a continuous bijection, and $X$ is compact and path-connected, and $Y$ is Hausdorff, then $f(X)subseteq Y$ is path-connected.
Now to answer your question directly. Let $X=mathbb{R}^2$ and $Y=mathbb{R}$ with the standard topology. Both of these spaces are path connected. We can construct a bijection between the two. Let $psi(x,y)=||x+y||-||x-y||$
this map is clearly surjective. It is also clearly injective. So it is bijective, and we know the norm function is continuous. So this is a continuous bijection, both of $mathbb{R}^2$ and $mathbb{R}$ are path-connected, but is well known that $X$ and $Y$ are not homeomorphic to each other.
Why don't we have a homemorphism? We have a continuous bijection, $X$ and $Y$ are Hausdorff, but $mathbb{R}^2$ is not compact. So the assumption of compactness is essential.
answered Oct 24 '16 at 2:29
Kernel_DirichletKernel_Dirichlet
1,138416
$endgroup$
1
$begingroup$
Your function is symmetric in both arguments so it is certainly not bijective
$endgroup$
– ThorbenK
Jan 8 at 17:01
add a comment |
$begingroup$
Yes, there are. But first lets consider criteria for homeomorphisms.
let $psi$$:Xrightarrow Y$ be a continuous bijection where $X$ is compact and $Y$ is Hausdorff. This is a homeomorphism. This means that topological invariants such as path-connectedness are preserved. So if we DO have a continuous bijection, and $X$ is compact and path-connected, and $Y$ is Hausdorff, then $f(X)subseteq Y$ is path-connected.
Now to answer your question directly. Let $X=mathbb{R}^2$ and $Y=mathbb{R}$ with the standard topology. Both of these spaces are path connected. We can construct a bijection between the two. Let $psi(x,y)=||x+y||-||x-y||$
this map is clearly surjective. It is also clearly injective. So it is bijective, and we know the norm function is continuous. So this is a continuous bijection, both of $mathbb{R}^2$ and $mathbb{R}$ are path-connected, but is well known that $X$ and $Y$ are not homeomorphic to each other.
Why don't we have a homemorphism? We have a continuous bijection, $X$ and $Y$ are Hausdorff, but $mathbb{R}^2$ is not compact. So the assumption of compactness is essential.
answered Oct 24 '16 at 2:29
Kernel_DirichletKernel_Dirichlet
1,138416
$endgroup$
Yes, there are. But first lets consider criteria for homeomorphisms.
let $psi$$:Xrightarrow Y$ be a continuous bijection where $X$ is compact and $Y$ is Hausdorff. This is a homeomorphism. This means that topological invariants such as path-connectedness are preserved. So if we DO have a continuous bijection, and $X$ is compact and path-connected, and $Y$ is Hausdorff, then $f(X)subseteq Y$ is path-connected.
Now to answer your question directly. Let $X=mathbb{R}^2$ and $Y=mathbb{R}$ with the standard topology. Both of these spaces are path connected. We can construct a bijection between the two. Let $psi(x,y)=||x+y||-||x-y||$
this map is clearly surjective. It is also clearly injective. So it is bijective, and we know the norm function is continuous. So this is a continuous bijection, both of $mathbb{R}^2$ and $mathbb{R}$ are path-connected, but is well known that $X$ and $Y$ are not homeomorphic to each other.
Why don't we have a homemorphism? We have a continuous bijection, $X$ and $Y$ are Hausdorff, but $mathbb{R}^2$ is not compact. So the assumption of compactness is essential.
answered Oct 24 '16 at 2:29
Kernel_DirichletKernel_Dirichlet
1,138416
edited Jan 8 at 16:20
answered Oct 24 '16 at 2:29
Kernel_DirichletKernel_Dirichlet
1,138416
answered Oct 24 '16 at 2:29
Kernel_DirichletKernel_Dirichlet
1,138416
answered Oct 24 '16 at 2:29
Kernel_DirichletKernel_Dirichlet
1,138416
1,138416
1
$begingroup$
Your function is symmetric in both arguments so it is certainly not bijective
$endgroup$
– ThorbenK
Jan 8 at 17:01
add a comment |
1
$begingroup$
Your function is symmetric in both arguments so it is certainly not bijective
$endgroup$
– ThorbenK
Jan 8 at 17:01
1
1
$begingroup$
Your function is symmetric in both arguments so it is certainly not bijective
$endgroup$
– ThorbenK
Jan 8 at 17:01
$begingroup$
Your function is symmetric in both arguments so it is certainly not bijective
$endgroup$
– ThorbenK
Jan 8 at 17:01
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%2f355763%2fpath-connectedness-and-continuous-bijections%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
$begingroup$
My mistake: I overlooked that you wanted a map in both directions. Please ignore my vote for closure.
$endgroup$
– Martin
Apr 9 '13 at 14:28