Bijections between Manifolds of the same dimension












0














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.










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  • 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


















0














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.










share|cite|improve this question






















  • 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
















0












0








0







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.










share|cite|improve this question













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






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asked Dec 4 '18 at 10:15









user39082

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




















  • 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












1 Answer
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oldest

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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.






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  • 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











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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.






share|cite|improve this answer























  • 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
















2














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.






share|cite|improve this answer























  • 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














2












2








2






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 16:54

























answered Dec 4 '18 at 18:38









Gustavo

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


















  • 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


















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