Proving that $[0,1] times X cong [0,1] times Y$, where $X$ is Möbius strip, $Y$ is curved surface of...
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I want to prove that $[0,1] times X cong [0,1] times Y$ where $[0,1] subset mathbb R$ has the usual Euclidean topology, $X$ is a Möbius strip and $Y$ is the curved surface of a cylinder. Here, $cong$ denotes that there exists a homeomorphism between the two spaces.
I know how to express $X$ and $Y$ as quotient spaces of $[0,1] times [0,1]$ but I'm stuck on what to do next.
general-topology mobius-band
edited Dec 15 '18 at 9:25
Brahadeesh
6,47942363
asked Jan 29 '18 at 18:58
user85798user85798
1
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|
show 5 more comments
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I want to prove that $[0,1] times X cong [0,1] times Y$ where $[0,1] subset mathbb R$ has the usual Euclidean topology, $X$ is a Möbius strip and $Y$ is the curved surface of a cylinder. Here, $cong$ denotes that there exists a homeomorphism between the two spaces.
I know how to express $X$ and $Y$ as quotient spaces of $[0,1] times [0,1]$ but I'm stuck on what to do next.
general-topology mobius-band
edited Dec 15 '18 at 9:25
Brahadeesh
6,47942363
asked Jan 29 '18 at 18:58
user85798user85798
1
$endgroup$
$begingroup$
What does the identity on $[0,1]times [0,1]$ look like under these quotients?
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– Fimpellizieri
Jan 29 '18 at 19:03
2
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Well, then you are out of luck since the first space is non-orientable and the second one is, which means that they cannot be homeomorphic.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 3:29
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@MoisheCohen Wait, are you sure? Can't you just rotate?
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– user85798
Jan 30 '18 at 13:14
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Rotate what? It is a general fact that the product of a non-orientable manifold with another manifold is again non-orientable.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 13:20
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@MoisheCohen I mean rotate the map continuously as you go around the strip, so that the normal vectors line up
$endgroup$
– user85798
Jan 30 '18 at 13:26
|
show 5 more comments
$begingroup$
I want to prove that $[0,1] times X cong [0,1] times Y$ where $[0,1] subset mathbb R$ has the usual Euclidean topology, $X$ is a Möbius strip and $Y$ is the curved surface of a cylinder. Here, $cong$ denotes that there exists a homeomorphism between the two spaces.
I know how to express $X$ and $Y$ as quotient spaces of $[0,1] times [0,1]$ but I'm stuck on what to do next.
general-topology mobius-band
edited Dec 15 '18 at 9:25
Brahadeesh
6,47942363
asked Jan 29 '18 at 18:58
user85798user85798
1
$endgroup$
I want to prove that $[0,1] times X cong [0,1] times Y$ where $[0,1] subset mathbb R$ has the usual Euclidean topology, $X$ is a Möbius strip and $Y$ is the curved surface of a cylinder. Here, $cong$ denotes that there exists a homeomorphism between the two spaces.
I know how to express $X$ and $Y$ as quotient spaces of $[0,1] times [0,1]$ but I'm stuck on what to do next.
general-topology mobius-band
general-topology mobius-band
edited Dec 15 '18 at 9:25
Brahadeesh
6,47942363
asked Jan 29 '18 at 18:58
user85798user85798
1
edited Dec 15 '18 at 9:25
Brahadeesh
6,47942363
asked Jan 29 '18 at 18:58
user85798user85798
1
edited Dec 15 '18 at 9:25
Brahadeesh
6,47942363
edited Dec 15 '18 at 9:25
Brahadeesh
6,47942363
edited Dec 15 '18 at 9:25
Brahadeesh
6,47942363
6,47942363
asked Jan 29 '18 at 18:58
user85798user85798
1
asked Jan 29 '18 at 18:58
user85798user85798
1
asked Jan 29 '18 at 18:58
user85798user85798
1
1
$begingroup$
What does the identity on $[0,1]times [0,1]$ look like under these quotients?
$endgroup$
– Fimpellizieri
Jan 29 '18 at 19:03
2
$begingroup$
Well, then you are out of luck since the first space is non-orientable and the second one is, which means that they cannot be homeomorphic.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 3:29
$begingroup$
@MoisheCohen Wait, are you sure? Can't you just rotate?
$endgroup$
– user85798
Jan 30 '18 at 13:14
$begingroup$
Rotate what? It is a general fact that the product of a non-orientable manifold with another manifold is again non-orientable.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 13:20
$begingroup$
@MoisheCohen I mean rotate the map continuously as you go around the strip, so that the normal vectors line up
$endgroup$
– user85798
Jan 30 '18 at 13:26
|
show 5 more comments
$begingroup$
What does the identity on $[0,1]times [0,1]$ look like under these quotients?
$endgroup$
– Fimpellizieri
Jan 29 '18 at 19:03
2
$begingroup$
Well, then you are out of luck since the first space is non-orientable and the second one is, which means that they cannot be homeomorphic.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 3:29
$begingroup$
@MoisheCohen Wait, are you sure? Can't you just rotate?
$endgroup$
– user85798
Jan 30 '18 at 13:14
$begingroup$
Rotate what? It is a general fact that the product of a non-orientable manifold with another manifold is again non-orientable.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 13:20
$begingroup$
@MoisheCohen I mean rotate the map continuously as you go around the strip, so that the normal vectors line up
$endgroup$
– user85798
Jan 30 '18 at 13:26
$begingroup$
What does the identity on $[0,1]times [0,1]$ look like under these quotients?
$endgroup$
– Fimpellizieri
Jan 29 '18 at 19:03
$begingroup$
What does the identity on $[0,1]times [0,1]$ look like under these quotients?
$endgroup$
– Fimpellizieri
Jan 29 '18 at 19:03
2
2
$begingroup$
Well, then you are out of luck since the first space is non-orientable and the second one is, which means that they cannot be homeomorphic.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 3:29
$begingroup$
Well, then you are out of luck since the first space is non-orientable and the second one is, which means that they cannot be homeomorphic.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 3:29
$begingroup$
@MoisheCohen Wait, are you sure? Can't you just rotate?
$endgroup$
– user85798
Jan 30 '18 at 13:14
$begingroup$
@MoisheCohen Wait, are you sure? Can't you just rotate?
$endgroup$
– user85798
Jan 30 '18 at 13:14
$begingroup$
Rotate what? It is a general fact that the product of a non-orientable manifold with another manifold is again non-orientable.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 13:20
$begingroup$
Rotate what? It is a general fact that the product of a non-orientable manifold with another manifold is again non-orientable.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 13:20
$begingroup$
@MoisheCohen I mean rotate the map continuously as you go around the strip, so that the normal vectors line up
$endgroup$
– user85798
Jan 30 '18 at 13:26
$begingroup$
@MoisheCohen I mean rotate the map continuously as you go around the strip, so that the normal vectors line up
$endgroup$
– user85798
Jan 30 '18 at 13:26
|
show 5 more comments
1 Answer
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From the comments above.
Orientability is preserved under homeomorphisms and it is a general fact that if $X$ is non-orientable and $Y$ and $Z$ are orientable, then $Z times X$ is non-orientable and $Z times Y$ is orientable. In your case, we have $Z = [0,1]$, $X = $ Möbius strip and $Y = $ curved surface of cylinder. So,
$$
[0,1] times X notcong [0,1] times Y.
$$
$endgroup$
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1 Answer
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$begingroup$
From the comments above.
Orientability is preserved under homeomorphisms and it is a general fact that if $X$ is non-orientable and $Y$ and $Z$ are orientable, then $Z times X$ is non-orientable and $Z times Y$ is orientable. In your case, we have $Z = [0,1]$, $X = $ Möbius strip and $Y = $ curved surface of cylinder. So,
$$
[0,1] times X notcong [0,1] times Y.
$$
$endgroup$
add a comment |
$begingroup$
From the comments above.
Orientability is preserved under homeomorphisms and it is a general fact that if $X$ is non-orientable and $Y$ and $Z$ are orientable, then $Z times X$ is non-orientable and $Z times Y$ is orientable. In your case, we have $Z = [0,1]$, $X = $ Möbius strip and $Y = $ curved surface of cylinder. So,
$$
[0,1] times X notcong [0,1] times Y.
$$
$endgroup$
add a comment |
$begingroup$
From the comments above.
Orientability is preserved under homeomorphisms and it is a general fact that if $X$ is non-orientable and $Y$ and $Z$ are orientable, then $Z times X$ is non-orientable and $Z times Y$ is orientable. In your case, we have $Z = [0,1]$, $X = $ Möbius strip and $Y = $ curved surface of cylinder. So,
$$
[0,1] times X notcong [0,1] times Y.
$$
$endgroup$
From the comments above.
Orientability is preserved under homeomorphisms and it is a general fact that if $X$ is non-orientable and $Y$ and $Z$ are orientable, then $Z times X$ is non-orientable and $Z times Y$ is orientable. In your case, we have $Z = [0,1]$, $X = $ Möbius strip and $Y = $ curved surface of cylinder. So,
$$
[0,1] times X notcong [0,1] times Y.
$$
answered Dec 15 '18 at 9:23
community wiki
Brahadeesh
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$begingroup$
What does the identity on $[0,1]times [0,1]$ look like under these quotients?
$endgroup$
– Fimpellizieri
Jan 29 '18 at 19:03
2
$begingroup$
Well, then you are out of luck since the first space is non-orientable and the second one is, which means that they cannot be homeomorphic.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 3:29
$begingroup$
@MoisheCohen Wait, are you sure? Can't you just rotate?
$endgroup$
– user85798
Jan 30 '18 at 13:14
$begingroup$
Rotate what? It is a general fact that the product of a non-orientable manifold with another manifold is again non-orientable.
$endgroup$
– Moishe Cohen
Jan 30 '18 at 13:20
$begingroup$
@MoisheCohen I mean rotate the map continuously as you go around the strip, so that the normal vectors line up
$endgroup$
– user85798
Jan 30 '18 at 13:26