Surface of a 2-sphere expressed as union of two closed disks
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I'm reading a First Course in Differential Geometry by Chuan-Chih Hsiung and on page 8 he says "A closed disk that is homeomorphic to $I^2$ [i.e. $Itimes I$, where $I = [a, b]$] is connected. The surface $S^2$ of a 2-sphere can be expressed as the union of two closed disks with nonempty intersection."
I'm not sure what he means by the second sentence. Am I supposed to imagine two disks being deformed into the two halves of the sphere (so the disks touch each other at their circumferences)? I don't understand what it means to express the spherical surface as a union of two disks.
general-topology differential-geometry
asked Nov 18 '12 at 6:44
thobansterthobanster
3791414
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add a comment |
$begingroup$
I'm reading a First Course in Differential Geometry by Chuan-Chih Hsiung and on page 8 he says "A closed disk that is homeomorphic to $I^2$ [i.e. $Itimes I$, where $I = [a, b]$] is connected. The surface $S^2$ of a 2-sphere can be expressed as the union of two closed disks with nonempty intersection."
I'm not sure what he means by the second sentence. Am I supposed to imagine two disks being deformed into the two halves of the sphere (so the disks touch each other at their circumferences)? I don't understand what it means to express the spherical surface as a union of two disks.
general-topology differential-geometry
asked Nov 18 '12 at 6:44
thobansterthobanster
3791414
$endgroup$
2
$begingroup$
Yes, that's exactly it. The northern and southern (closed) hemispheres of a sphere are topologically (closed) disks, intersecting on their boundaries.
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– Robert Israel
Nov 18 '12 at 6:47
$begingroup$
And you can visualize their common boundary as the equator of the sphere.
$endgroup$
– Brian M. Scott
Nov 18 '12 at 7:28
add a comment |
$begingroup$
I'm reading a First Course in Differential Geometry by Chuan-Chih Hsiung and on page 8 he says "A closed disk that is homeomorphic to $I^2$ [i.e. $Itimes I$, where $I = [a, b]$] is connected. The surface $S^2$ of a 2-sphere can be expressed as the union of two closed disks with nonempty intersection."
I'm not sure what he means by the second sentence. Am I supposed to imagine two disks being deformed into the two halves of the sphere (so the disks touch each other at their circumferences)? I don't understand what it means to express the spherical surface as a union of two disks.
general-topology differential-geometry
asked Nov 18 '12 at 6:44
thobansterthobanster
3791414
$endgroup$
I'm reading a First Course in Differential Geometry by Chuan-Chih Hsiung and on page 8 he says "A closed disk that is homeomorphic to $I^2$ [i.e. $Itimes I$, where $I = [a, b]$] is connected. The surface $S^2$ of a 2-sphere can be expressed as the union of two closed disks with nonempty intersection."
I'm not sure what he means by the second sentence. Am I supposed to imagine two disks being deformed into the two halves of the sphere (so the disks touch each other at their circumferences)? I don't understand what it means to express the spherical surface as a union of two disks.
general-topology differential-geometry
general-topology differential-geometry
asked Nov 18 '12 at 6:44
thobansterthobanster
3791414
asked Nov 18 '12 at 6:44
thobansterthobanster
3791414
asked Nov 18 '12 at 6:44
thobansterthobanster
3791414
asked Nov 18 '12 at 6:44
thobansterthobanster
3791414
asked Nov 18 '12 at 6:44
thobansterthobanster
3791414
3791414
2
$begingroup$
Yes, that's exactly it. The northern and southern (closed) hemispheres of a sphere are topologically (closed) disks, intersecting on their boundaries.
$endgroup$
– Robert Israel
Nov 18 '12 at 6:47
$begingroup$
And you can visualize their common boundary as the equator of the sphere.
$endgroup$
– Brian M. Scott
Nov 18 '12 at 7:28
add a comment |
2
$begingroup$
Yes, that's exactly it. The northern and southern (closed) hemispheres of a sphere are topologically (closed) disks, intersecting on their boundaries.
$endgroup$
– Robert Israel
Nov 18 '12 at 6:47
$begingroup$
And you can visualize their common boundary as the equator of the sphere.
$endgroup$
– Brian M. Scott
Nov 18 '12 at 7:28
2
2
$begingroup$
Yes, that's exactly it. The northern and southern (closed) hemispheres of a sphere are topologically (closed) disks, intersecting on their boundaries.
$endgroup$
– Robert Israel
Nov 18 '12 at 6:47
$begingroup$
Yes, that's exactly it. The northern and southern (closed) hemispheres of a sphere are topologically (closed) disks, intersecting on their boundaries.
$endgroup$
– Robert Israel
Nov 18 '12 at 6:47
$begingroup$
And you can visualize their common boundary as the equator of the sphere.
$endgroup$
– Brian M. Scott
Nov 18 '12 at 7:28
$begingroup$
And you can visualize their common boundary as the equator of the sphere.
$endgroup$
– Brian M. Scott
Nov 18 '12 at 7:28
add a comment |
1 Answer
1
active
oldest
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Intuitively, you are gluing the two disks along their boundaries.
This can be made precise with the language of pushouts in $textbf{Top}$ : Given $f : X to Y$, $G : X to Z$, the pushout of $f$ and $g$, denoted by $Y times_X Z$ is defined to be $Y sqcup Z / sim$ where $y sim z$ provided there exists $x in X$ such that $f(x) = y$ and $g(x) = z$. Intuitively, you are gluing along the information given by $f$ and $g$.
In this particular situation, we have the following,
$$
require{AMScd} begin{CD}
S^1 @>{operatorname{inc}}>> D^2\ @V{operatorname{inc}}VV @VV{}V\
D^2 @>>{}> S^2
end{CD}
$$
Visually,
You can easily see that once you have the upper hemisphere (or lower) you can project onto the $xy$-plane by forgetting the $z$-coordinate. This is a homeomorphism, so taking the inverse allows you to replace $D^2$ with this "curved" disk, coinciding with the upper (or lower) hemisphere.
answered Dec 10 '18 at 20:32
Robert CardonaRobert Cardona
5,382234102
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add a comment |
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$begingroup$
Intuitively, you are gluing the two disks along their boundaries.
This can be made precise with the language of pushouts in $textbf{Top}$ : Given $f : X to Y$, $G : X to Z$, the pushout of $f$ and $g$, denoted by $Y times_X Z$ is defined to be $Y sqcup Z / sim$ where $y sim z$ provided there exists $x in X$ such that $f(x) = y$ and $g(x) = z$. Intuitively, you are gluing along the information given by $f$ and $g$.
In this particular situation, we have the following,
$$
require{AMScd} begin{CD}
S^1 @>{operatorname{inc}}>> D^2\ @V{operatorname{inc}}VV @VV{}V\
D^2 @>>{}> S^2
end{CD}
$$
Visually,
You can easily see that once you have the upper hemisphere (or lower) you can project onto the $xy$-plane by forgetting the $z$-coordinate. This is a homeomorphism, so taking the inverse allows you to replace $D^2$ with this "curved" disk, coinciding with the upper (or lower) hemisphere.
answered Dec 10 '18 at 20:32
Robert CardonaRobert Cardona
5,382234102
$endgroup$
add a comment |
$begingroup$
Intuitively, you are gluing the two disks along their boundaries.
This can be made precise with the language of pushouts in $textbf{Top}$ : Given $f : X to Y$, $G : X to Z$, the pushout of $f$ and $g$, denoted by $Y times_X Z$ is defined to be $Y sqcup Z / sim$ where $y sim z$ provided there exists $x in X$ such that $f(x) = y$ and $g(x) = z$. Intuitively, you are gluing along the information given by $f$ and $g$.
In this particular situation, we have the following,
$$
require{AMScd} begin{CD}
S^1 @>{operatorname{inc}}>> D^2\ @V{operatorname{inc}}VV @VV{}V\
D^2 @>>{}> S^2
end{CD}
$$
Visually,
You can easily see that once you have the upper hemisphere (or lower) you can project onto the $xy$-plane by forgetting the $z$-coordinate. This is a homeomorphism, so taking the inverse allows you to replace $D^2$ with this "curved" disk, coinciding with the upper (or lower) hemisphere.
answered Dec 10 '18 at 20:32
Robert CardonaRobert Cardona
5,382234102
$endgroup$
add a comment |
$begingroup$
Intuitively, you are gluing the two disks along their boundaries.
This can be made precise with the language of pushouts in $textbf{Top}$ : Given $f : X to Y$, $G : X to Z$, the pushout of $f$ and $g$, denoted by $Y times_X Z$ is defined to be $Y sqcup Z / sim$ where $y sim z$ provided there exists $x in X$ such that $f(x) = y$ and $g(x) = z$. Intuitively, you are gluing along the information given by $f$ and $g$.
In this particular situation, we have the following,
$$
require{AMScd} begin{CD}
S^1 @>{operatorname{inc}}>> D^2\ @V{operatorname{inc}}VV @VV{}V\
D^2 @>>{}> S^2
end{CD}
$$
Visually,
You can easily see that once you have the upper hemisphere (or lower) you can project onto the $xy$-plane by forgetting the $z$-coordinate. This is a homeomorphism, so taking the inverse allows you to replace $D^2$ with this "curved" disk, coinciding with the upper (or lower) hemisphere.
answered Dec 10 '18 at 20:32
Robert CardonaRobert Cardona
5,382234102
$endgroup$
Intuitively, you are gluing the two disks along their boundaries.
This can be made precise with the language of pushouts in $textbf{Top}$ : Given $f : X to Y$, $G : X to Z$, the pushout of $f$ and $g$, denoted by $Y times_X Z$ is defined to be $Y sqcup Z / sim$ where $y sim z$ provided there exists $x in X$ such that $f(x) = y$ and $g(x) = z$. Intuitively, you are gluing along the information given by $f$ and $g$.
In this particular situation, we have the following,
$$
require{AMScd} begin{CD}
S^1 @>{operatorname{inc}}>> D^2\ @V{operatorname{inc}}VV @VV{}V\
D^2 @>>{}> S^2
end{CD}
$$
Visually,
You can easily see that once you have the upper hemisphere (or lower) you can project onto the $xy$-plane by forgetting the $z$-coordinate. This is a homeomorphism, so taking the inverse allows you to replace $D^2$ with this "curved" disk, coinciding with the upper (or lower) hemisphere.
answered Dec 10 '18 at 20:32
Robert CardonaRobert Cardona
5,382234102
answered Dec 10 '18 at 20:32
Robert CardonaRobert Cardona
5,382234102
answered Dec 10 '18 at 20:32
Robert CardonaRobert Cardona
5,382234102
answered Dec 10 '18 at 20:32
Robert CardonaRobert Cardona
5,382234102
5,382234102
add a comment |
add a comment |
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$begingroup$
Yes, that's exactly it. The northern and southern (closed) hemispheres of a sphere are topologically (closed) disks, intersecting on their boundaries.
$endgroup$
– Robert Israel
Nov 18 '12 at 6:47
$begingroup$
And you can visualize their common boundary as the equator of the sphere.
$endgroup$
– Brian M. Scott
Nov 18 '12 at 7:28