Bijections between Manifolds of the same dimension
Given a continuous bijection between manifolds of the same dimension, does it have to be a homeomorphism?
I know that this has a straightforward proof for compact Hausdorff space, be they manifolds or not.
I also know that there are continuous bijections from non-compact manifolds to subsets of higher-dimensional manifolds, the easiest example being non-periodic curves on the torus as images of the real line with dense image. In that case the continuous bijection is not homeomorphism, but the subspace topology of the image is not that of a manifold anyway.
I think that this should be true for the following reason. Since continuity of the inverse map is a local property, the question should reduce (in chart neighborhoods of an arbitrary point and its image) to showing that a continuous bijection between n-balls is a homeomorphism. But this is true because the n-ball is a compact Hausdorff space. Anyway I would prefer to have a citeable reference.
general-topology algebraic-topology continuity manifolds
add a comment |
Given a continuous bijection between manifolds of the same dimension, does it have to be a homeomorphism?
I know that this has a straightforward proof for compact Hausdorff space, be they manifolds or not.
I also know that there are continuous bijections from non-compact manifolds to subsets of higher-dimensional manifolds, the easiest example being non-periodic curves on the torus as images of the real line with dense image. In that case the continuous bijection is not homeomorphism, but the subspace topology of the image is not that of a manifold anyway.
I think that this should be true for the following reason. Since continuity of the inverse map is a local property, the question should reduce (in chart neighborhoods of an arbitrary point and its image) to showing that a continuous bijection between n-balls is a homeomorphism. But this is true because the n-ball is a compact Hausdorff space. Anyway I would prefer to have a citeable reference.
general-topology algebraic-topology continuity manifolds
Doesn't this boil down to the question whether the inverse of a bijective continuous function is again continuous? Which is false.
– James
Dec 4 '18 at 10:17
1
This is a direct consequence of the invariance of domain theorem. Can you produce a simple proof using that? en.wikipedia.org/wiki/Invariance_of_domain
– Gustavo
Dec 4 '18 at 10:29
I see. This was what I was looking for. If you write it as an answer, I will accept it.
– user39082
Dec 4 '18 at 11:08
1
If you consider manifolds with boundary, then it is not true. The map $e : ´[0,2pi) to S^1, e(t) = e^{it}$, is a continuous bijection.
– Paul Frost
Dec 4 '18 at 18:05
add a comment |
Given a continuous bijection between manifolds of the same dimension, does it have to be a homeomorphism?
I know that this has a straightforward proof for compact Hausdorff space, be they manifolds or not.
I also know that there are continuous bijections from non-compact manifolds to subsets of higher-dimensional manifolds, the easiest example being non-periodic curves on the torus as images of the real line with dense image. In that case the continuous bijection is not homeomorphism, but the subspace topology of the image is not that of a manifold anyway.
I think that this should be true for the following reason. Since continuity of the inverse map is a local property, the question should reduce (in chart neighborhoods of an arbitrary point and its image) to showing that a continuous bijection between n-balls is a homeomorphism. But this is true because the n-ball is a compact Hausdorff space. Anyway I would prefer to have a citeable reference.
general-topology algebraic-topology continuity manifolds
Given a continuous bijection between manifolds of the same dimension, does it have to be a homeomorphism?
I know that this has a straightforward proof for compact Hausdorff space, be they manifolds or not.
I also know that there are continuous bijections from non-compact manifolds to subsets of higher-dimensional manifolds, the easiest example being non-periodic curves on the torus as images of the real line with dense image. In that case the continuous bijection is not homeomorphism, but the subspace topology of the image is not that of a manifold anyway.
I think that this should be true for the following reason. Since continuity of the inverse map is a local property, the question should reduce (in chart neighborhoods of an arbitrary point and its image) to showing that a continuous bijection between n-balls is a homeomorphism. But this is true because the n-ball is a compact Hausdorff space. Anyway I would prefer to have a citeable reference.
general-topology algebraic-topology continuity manifolds
general-topology algebraic-topology continuity manifolds
asked Dec 4 '18 at 10:15
user39082user39082
1,235513
1,235513
Doesn't this boil down to the question whether the inverse of a bijective continuous function is again continuous? Which is false.
– James
Dec 4 '18 at 10:17
1
This is a direct consequence of the invariance of domain theorem. Can you produce a simple proof using that? en.wikipedia.org/wiki/Invariance_of_domain
– Gustavo
Dec 4 '18 at 10:29
I see. This was what I was looking for. If you write it as an answer, I will accept it.
– user39082
Dec 4 '18 at 11:08
1
If you consider manifolds with boundary, then it is not true. The map $e : ´[0,2pi) to S^1, e(t) = e^{it}$, is a continuous bijection.
– Paul Frost
Dec 4 '18 at 18:05
add a comment |
Doesn't this boil down to the question whether the inverse of a bijective continuous function is again continuous? Which is false.
– James
Dec 4 '18 at 10:17
1
This is a direct consequence of the invariance of domain theorem. Can you produce a simple proof using that? en.wikipedia.org/wiki/Invariance_of_domain
– Gustavo
Dec 4 '18 at 10:29
I see. This was what I was looking for. If you write it as an answer, I will accept it.
– user39082
Dec 4 '18 at 11:08
1
If you consider manifolds with boundary, then it is not true. The map $e : ´[0,2pi) to S^1, e(t) = e^{it}$, is a continuous bijection.
– Paul Frost
Dec 4 '18 at 18:05
Doesn't this boil down to the question whether the inverse of a bijective continuous function is again continuous? Which is false.
– James
Dec 4 '18 at 10:17
Doesn't this boil down to the question whether the inverse of a bijective continuous function is again continuous? Which is false.
– James
Dec 4 '18 at 10:17
1
1
This is a direct consequence of the invariance of domain theorem. Can you produce a simple proof using that? en.wikipedia.org/wiki/Invariance_of_domain
– Gustavo
Dec 4 '18 at 10:29
This is a direct consequence of the invariance of domain theorem. Can you produce a simple proof using that? en.wikipedia.org/wiki/Invariance_of_domain
– Gustavo
Dec 4 '18 at 10:29
I see. This was what I was looking for. If you write it as an answer, I will accept it.
– user39082
Dec 4 '18 at 11:08
I see. This was what I was looking for. If you write it as an answer, I will accept it.
– user39082
Dec 4 '18 at 11:08
1
1
If you consider manifolds with boundary, then it is not true. The map $e : ´[0,2pi) to S^1, e(t) = e^{it}$, is a continuous bijection.
– Paul Frost
Dec 4 '18 at 18:05
If you consider manifolds with boundary, then it is not true. The map $e : ´[0,2pi) to S^1, e(t) = e^{it}$, is a continuous bijection.
– Paul Frost
Dec 4 '18 at 18:05
add a comment |
1 Answer
1
active
oldest
votes
Let $X$ and $Y$ be $m$-manifolds (without boundary) and $ f:X to Y $ be a continuous bijection. Let $ p in X $. By definition there exists open sets $ A subset X $ and $ B subset Y $ and charts $ x: A to mathbb{R}^m $ and $ y: B to mathbb{R}^m $ such that $ p in A $ and $ f(p) in B $. Let $ f_{xy} : im(x) to mathbb{R}^m $ be such that $ f_{xy} (q) = y Big( f big( x^{-1} (q) big) Big) $, $forall q in im(x) $. That is $ f_{xy} = (y|_{f[A]}) circ (f|_A) circ x^{-1} $. Now $ im(x) subset mathbb{R}^m $ is open (since $ x $ is a homeomorphism) and $ f_{xy} $ is injective continuous (since $ x^{-1}$, $f|_A $ and $ y|_{f[A]} $ are injective continuous). By invariance of domain we have that $ f_{xy}[im(x)] = im(f_{xy}) = y big[ f[A] big] $ is open and $ f_{xy} $ is a homeomorphism between $ im(x) $ and $ y big[ f[A] big] $. Therefore $ D = B cap f[A] = y^{-1} Big[ y big[ f[A] big] Big] $ and $ C = A cap f^{-1} [B] = f^{-1} big[ B cap f[A] big] $ are open, $ x[C] = f_{xy}^{-1} big[ y[D] big] $, $ y big[ f[A] big] = y[D] $, $ y|_{f[A]} = y|_D $ and $ p in C $. We are left with a continuous bijection $ f|_C : C to D $ whose inverse is $ f|_C^{-1} = big( x^{-1}|_{x[C]} big) circ big( f_{xy}^{-1}|_{y[D]} big) circ (y|_D) : D to C $. Thus $ f|_C^{-1} $ is continuous because $ x^{-1}|_{x[C]} , $, $f_{xy}^{-1}|_{y[D]} $ and $ y|_D $ are continuous. Thus $ p in C$, $C subset X $ is open, $D subset Y $ is open and $ f|_C :C to D $ is a homeomorphism. Since $p$ is arbitrary, $f: X to Y $ is a local homeomorphism and since bijective local homeomorphisms are global homeomorphisms, we are done.
This answer does not make any sense to me — half-open interval have obvious bijective map to a circle, and it is clearly not a homeomorphism.
– xsnl
Dec 5 '18 at 15:01
2
The answer covers the case of manifolds without boundary. I edited to make it clear.
– Gustavo
Dec 5 '18 at 16:56
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%2f3025388%2fbijections-between-manifolds-of-the-same-dimension%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
Let $X$ and $Y$ be $m$-manifolds (without boundary) and $ f:X to Y $ be a continuous bijection. Let $ p in X $. By definition there exists open sets $ A subset X $ and $ B subset Y $ and charts $ x: A to mathbb{R}^m $ and $ y: B to mathbb{R}^m $ such that $ p in A $ and $ f(p) in B $. Let $ f_{xy} : im(x) to mathbb{R}^m $ be such that $ f_{xy} (q) = y Big( f big( x^{-1} (q) big) Big) $, $forall q in im(x) $. That is $ f_{xy} = (y|_{f[A]}) circ (f|_A) circ x^{-1} $. Now $ im(x) subset mathbb{R}^m $ is open (since $ x $ is a homeomorphism) and $ f_{xy} $ is injective continuous (since $ x^{-1}$, $f|_A $ and $ y|_{f[A]} $ are injective continuous). By invariance of domain we have that $ f_{xy}[im(x)] = im(f_{xy}) = y big[ f[A] big] $ is open and $ f_{xy} $ is a homeomorphism between $ im(x) $ and $ y big[ f[A] big] $. Therefore $ D = B cap f[A] = y^{-1} Big[ y big[ f[A] big] Big] $ and $ C = A cap f^{-1} [B] = f^{-1} big[ B cap f[A] big] $ are open, $ x[C] = f_{xy}^{-1} big[ y[D] big] $, $ y big[ f[A] big] = y[D] $, $ y|_{f[A]} = y|_D $ and $ p in C $. We are left with a continuous bijection $ f|_C : C to D $ whose inverse is $ f|_C^{-1} = big( x^{-1}|_{x[C]} big) circ big( f_{xy}^{-1}|_{y[D]} big) circ (y|_D) : D to C $. Thus $ f|_C^{-1} $ is continuous because $ x^{-1}|_{x[C]} , $, $f_{xy}^{-1}|_{y[D]} $ and $ y|_D $ are continuous. Thus $ p in C$, $C subset X $ is open, $D subset Y $ is open and $ f|_C :C to D $ is a homeomorphism. Since $p$ is arbitrary, $f: X to Y $ is a local homeomorphism and since bijective local homeomorphisms are global homeomorphisms, we are done.
This answer does not make any sense to me — half-open interval have obvious bijective map to a circle, and it is clearly not a homeomorphism.
– xsnl
Dec 5 '18 at 15:01
2
The answer covers the case of manifolds without boundary. I edited to make it clear.
– Gustavo
Dec 5 '18 at 16:56
add a comment |
Let $X$ and $Y$ be $m$-manifolds (without boundary) and $ f:X to Y $ be a continuous bijection. Let $ p in X $. By definition there exists open sets $ A subset X $ and $ B subset Y $ and charts $ x: A to mathbb{R}^m $ and $ y: B to mathbb{R}^m $ such that $ p in A $ and $ f(p) in B $. Let $ f_{xy} : im(x) to mathbb{R}^m $ be such that $ f_{xy} (q) = y Big( f big( x^{-1} (q) big) Big) $, $forall q in im(x) $. That is $ f_{xy} = (y|_{f[A]}) circ (f|_A) circ x^{-1} $. Now $ im(x) subset mathbb{R}^m $ is open (since $ x $ is a homeomorphism) and $ f_{xy} $ is injective continuous (since $ x^{-1}$, $f|_A $ and $ y|_{f[A]} $ are injective continuous). By invariance of domain we have that $ f_{xy}[im(x)] = im(f_{xy}) = y big[ f[A] big] $ is open and $ f_{xy} $ is a homeomorphism between $ im(x) $ and $ y big[ f[A] big] $. Therefore $ D = B cap f[A] = y^{-1} Big[ y big[ f[A] big] Big] $ and $ C = A cap f^{-1} [B] = f^{-1} big[ B cap f[A] big] $ are open, $ x[C] = f_{xy}^{-1} big[ y[D] big] $, $ y big[ f[A] big] = y[D] $, $ y|_{f[A]} = y|_D $ and $ p in C $. We are left with a continuous bijection $ f|_C : C to D $ whose inverse is $ f|_C^{-1} = big( x^{-1}|_{x[C]} big) circ big( f_{xy}^{-1}|_{y[D]} big) circ (y|_D) : D to C $. Thus $ f|_C^{-1} $ is continuous because $ x^{-1}|_{x[C]} , $, $f_{xy}^{-1}|_{y[D]} $ and $ y|_D $ are continuous. Thus $ p in C$, $C subset X $ is open, $D subset Y $ is open and $ f|_C :C to D $ is a homeomorphism. Since $p$ is arbitrary, $f: X to Y $ is a local homeomorphism and since bijective local homeomorphisms are global homeomorphisms, we are done.
This answer does not make any sense to me — half-open interval have obvious bijective map to a circle, and it is clearly not a homeomorphism.
– xsnl
Dec 5 '18 at 15:01
2
The answer covers the case of manifolds without boundary. I edited to make it clear.
– Gustavo
Dec 5 '18 at 16:56
add a comment |
Let $X$ and $Y$ be $m$-manifolds (without boundary) and $ f:X to Y $ be a continuous bijection. Let $ p in X $. By definition there exists open sets $ A subset X $ and $ B subset Y $ and charts $ x: A to mathbb{R}^m $ and $ y: B to mathbb{R}^m $ such that $ p in A $ and $ f(p) in B $. Let $ f_{xy} : im(x) to mathbb{R}^m $ be such that $ f_{xy} (q) = y Big( f big( x^{-1} (q) big) Big) $, $forall q in im(x) $. That is $ f_{xy} = (y|_{f[A]}) circ (f|_A) circ x^{-1} $. Now $ im(x) subset mathbb{R}^m $ is open (since $ x $ is a homeomorphism) and $ f_{xy} $ is injective continuous (since $ x^{-1}$, $f|_A $ and $ y|_{f[A]} $ are injective continuous). By invariance of domain we have that $ f_{xy}[im(x)] = im(f_{xy}) = y big[ f[A] big] $ is open and $ f_{xy} $ is a homeomorphism between $ im(x) $ and $ y big[ f[A] big] $. Therefore $ D = B cap f[A] = y^{-1} Big[ y big[ f[A] big] Big] $ and $ C = A cap f^{-1} [B] = f^{-1} big[ B cap f[A] big] $ are open, $ x[C] = f_{xy}^{-1} big[ y[D] big] $, $ y big[ f[A] big] = y[D] $, $ y|_{f[A]} = y|_D $ and $ p in C $. We are left with a continuous bijection $ f|_C : C to D $ whose inverse is $ f|_C^{-1} = big( x^{-1}|_{x[C]} big) circ big( f_{xy}^{-1}|_{y[D]} big) circ (y|_D) : D to C $. Thus $ f|_C^{-1} $ is continuous because $ x^{-1}|_{x[C]} , $, $f_{xy}^{-1}|_{y[D]} $ and $ y|_D $ are continuous. Thus $ p in C$, $C subset X $ is open, $D subset Y $ is open and $ f|_C :C to D $ is a homeomorphism. Since $p$ is arbitrary, $f: X to Y $ is a local homeomorphism and since bijective local homeomorphisms are global homeomorphisms, we are done.
Let $X$ and $Y$ be $m$-manifolds (without boundary) and $ f:X to Y $ be a continuous bijection. Let $ p in X $. By definition there exists open sets $ A subset X $ and $ B subset Y $ and charts $ x: A to mathbb{R}^m $ and $ y: B to mathbb{R}^m $ such that $ p in A $ and $ f(p) in B $. Let $ f_{xy} : im(x) to mathbb{R}^m $ be such that $ f_{xy} (q) = y Big( f big( x^{-1} (q) big) Big) $, $forall q in im(x) $. That is $ f_{xy} = (y|_{f[A]}) circ (f|_A) circ x^{-1} $. Now $ im(x) subset mathbb{R}^m $ is open (since $ x $ is a homeomorphism) and $ f_{xy} $ is injective continuous (since $ x^{-1}$, $f|_A $ and $ y|_{f[A]} $ are injective continuous). By invariance of domain we have that $ f_{xy}[im(x)] = im(f_{xy}) = y big[ f[A] big] $ is open and $ f_{xy} $ is a homeomorphism between $ im(x) $ and $ y big[ f[A] big] $. Therefore $ D = B cap f[A] = y^{-1} Big[ y big[ f[A] big] Big] $ and $ C = A cap f^{-1} [B] = f^{-1} big[ B cap f[A] big] $ are open, $ x[C] = f_{xy}^{-1} big[ y[D] big] $, $ y big[ f[A] big] = y[D] $, $ y|_{f[A]} = y|_D $ and $ p in C $. We are left with a continuous bijection $ f|_C : C to D $ whose inverse is $ f|_C^{-1} = big( x^{-1}|_{x[C]} big) circ big( f_{xy}^{-1}|_{y[D]} big) circ (y|_D) : D to C $. Thus $ f|_C^{-1} $ is continuous because $ x^{-1}|_{x[C]} , $, $f_{xy}^{-1}|_{y[D]} $ and $ y|_D $ are continuous. Thus $ p in C$, $C subset X $ is open, $D subset Y $ is open and $ f|_C :C to D $ is a homeomorphism. Since $p$ is arbitrary, $f: X to Y $ is a local homeomorphism and since bijective local homeomorphisms are global homeomorphisms, we are done.
edited Dec 5 '18 at 16:54
answered Dec 4 '18 at 18:38
GustavoGustavo
866621
866621
This answer does not make any sense to me — half-open interval have obvious bijective map to a circle, and it is clearly not a homeomorphism.
– xsnl
Dec 5 '18 at 15:01
2
The answer covers the case of manifolds without boundary. I edited to make it clear.
– Gustavo
Dec 5 '18 at 16:56
add a comment |
This answer does not make any sense to me — half-open interval have obvious bijective map to a circle, and it is clearly not a homeomorphism.
– xsnl
Dec 5 '18 at 15:01
2
The answer covers the case of manifolds without boundary. I edited to make it clear.
– Gustavo
Dec 5 '18 at 16:56
This answer does not make any sense to me — half-open interval have obvious bijective map to a circle, and it is clearly not a homeomorphism.
– xsnl
Dec 5 '18 at 15:01
This answer does not make any sense to me — half-open interval have obvious bijective map to a circle, and it is clearly not a homeomorphism.
– xsnl
Dec 5 '18 at 15:01
2
2
The answer covers the case of manifolds without boundary. I edited to make it clear.
– Gustavo
Dec 5 '18 at 16:56
The answer covers the case of manifolds without boundary. I edited to make it clear.
– Gustavo
Dec 5 '18 at 16:56
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%2f3025388%2fbijections-between-manifolds-of-the-same-dimension%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
Doesn't this boil down to the question whether the inverse of a bijective continuous function is again continuous? Which is false.
– James
Dec 4 '18 at 10:17
1
This is a direct consequence of the invariance of domain theorem. Can you produce a simple proof using that? en.wikipedia.org/wiki/Invariance_of_domain
– Gustavo
Dec 4 '18 at 10:29
I see. This was what I was looking for. If you write it as an answer, I will accept it.
– user39082
Dec 4 '18 at 11:08
1
If you consider manifolds with boundary, then it is not true. The map $e : ´[0,2pi) to S^1, e(t) = e^{it}$, is a continuous bijection.
– Paul Frost
Dec 4 '18 at 18:05