homotopy equivalence of these three spaces
$begingroup$
$S^2$ with a diameter
$T^2$ with a disk
$S^2$ with a circle
i try to find a space such that they are all the deformation contraction of it. And i failed.
any idea is helpful. thanks
algebraic-topology homotopy-theory
edited Dec 15 '18 at 11:01
Paul Frost
10.7k3934
asked Dec 15 '18 at 4:42
yufeng luyufeng lu
353
$endgroup$
add a comment |
$begingroup$
$S^2$ with a diameter
$T^2$ with a disk
$S^2$ with a circle
i try to find a space such that they are all the deformation contraction of it. And i failed.
any idea is helpful. thanks
algebraic-topology homotopy-theory
edited Dec 15 '18 at 11:01
Paul Frost
10.7k3934
asked Dec 15 '18 at 4:42
yufeng luyufeng lu
353
$endgroup$
$begingroup$
In $(a)$, contract the diameter and squeeze out a disc around the double point to go $(a)Rightarrow (b)$. Also in $(a)$, take one end of the diameter and drag it continuously around the surface of $S^2$ till it coincides with the point at the other end. This gives you $(c)$ with the $S^1$ inside the $S^2$, and this is completely fine.
$endgroup$
– Tyrone
Dec 15 '18 at 10:49
$begingroup$
Similar to math.stackexchange.com/q/3023209.
$endgroup$
– Paul Frost
Dec 15 '18 at 11:00
add a comment |
$begingroup$
$S^2$ with a diameter
$T^2$ with a disk
$S^2$ with a circle
i try to find a space such that they are all the deformation contraction of it. And i failed.
any idea is helpful. thanks
algebraic-topology homotopy-theory
edited Dec 15 '18 at 11:01
Paul Frost
10.7k3934
asked Dec 15 '18 at 4:42
yufeng luyufeng lu
353
$endgroup$
$S^2$ with a diameter
$T^2$ with a disk
$S^2$ with a circle
i try to find a space such that they are all the deformation contraction of it. And i failed.
any idea is helpful. thanks
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
edited Dec 15 '18 at 11:01
Paul Frost
10.7k3934
asked Dec 15 '18 at 4:42
yufeng luyufeng lu
353
edited Dec 15 '18 at 11:01
Paul Frost
10.7k3934
asked Dec 15 '18 at 4:42
yufeng luyufeng lu
353
edited Dec 15 '18 at 11:01
Paul Frost
10.7k3934
edited Dec 15 '18 at 11:01
Paul Frost
10.7k3934
edited Dec 15 '18 at 11:01
Paul Frost
10.7k3934
10.7k3934
asked Dec 15 '18 at 4:42
yufeng luyufeng lu
353
asked Dec 15 '18 at 4:42
yufeng luyufeng lu
353
asked Dec 15 '18 at 4:42
yufeng luyufeng lu
353
353
$begingroup$
In $(a)$, contract the diameter and squeeze out a disc around the double point to go $(a)Rightarrow (b)$. Also in $(a)$, take one end of the diameter and drag it continuously around the surface of $S^2$ till it coincides with the point at the other end. This gives you $(c)$ with the $S^1$ inside the $S^2$, and this is completely fine.
$endgroup$
– Tyrone
Dec 15 '18 at 10:49
$begingroup$
Similar to math.stackexchange.com/q/3023209.
$endgroup$
– Paul Frost
Dec 15 '18 at 11:00
add a comment |
$begingroup$
In $(a)$, contract the diameter and squeeze out a disc around the double point to go $(a)Rightarrow (b)$. Also in $(a)$, take one end of the diameter and drag it continuously around the surface of $S^2$ till it coincides with the point at the other end. This gives you $(c)$ with the $S^1$ inside the $S^2$, and this is completely fine.
$endgroup$
– Tyrone
Dec 15 '18 at 10:49
$begingroup$
Similar to math.stackexchange.com/q/3023209.
$endgroup$
– Paul Frost
Dec 15 '18 at 11:00
$begingroup$
In $(a)$, contract the diameter and squeeze out a disc around the double point to go $(a)Rightarrow (b)$. Also in $(a)$, take one end of the diameter and drag it continuously around the surface of $S^2$ till it coincides with the point at the other end. This gives you $(c)$ with the $S^1$ inside the $S^2$, and this is completely fine.
$endgroup$
– Tyrone
Dec 15 '18 at 10:49
$begingroup$
In $(a)$, contract the diameter and squeeze out a disc around the double point to go $(a)Rightarrow (b)$. Also in $(a)$, take one end of the diameter and drag it continuously around the surface of $S^2$ till it coincides with the point at the other end. This gives you $(c)$ with the $S^1$ inside the $S^2$, and this is completely fine.
$endgroup$
– Tyrone
Dec 15 '18 at 10:49
$begingroup$
Similar to math.stackexchange.com/q/3023209.
$endgroup$
– Paul Frost
Dec 15 '18 at 11:00
$begingroup$
Similar to math.stackexchange.com/q/3023209.
$endgroup$
– Paul Frost
Dec 15 '18 at 11:00
add a comment |
1 Answer
1
active
oldest
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$begingroup$
You want to show that all three spaces $S_1, S_2, S_3$ are homotopy equivalent. To do this, it is not necessary to find a space $S$ such that each $S_i$ is a deformation retract (but I shall come back to this point later).
I shall use the following two well-known theorems:
If $A hookrightarrow X$ is a cofibration and $A$ is contractible, then the qotient map $X to X/A$ is a homotopy equivalence.
If $X$ is a CW-complex and $A$ is a subcomplex, then $A hookrightarrow X$ is a cofibration.
Appyling these theorems to $(S_1,diameter)$ and $(S_2,disk)$, we see that both spaces are homotopy equivalent to $S^2/S^0$, where $S^0$ is regarded as the set consisting of north and south pole. That this space is homotopy equivalent to $S_3$ has been shown in Show $mathbb{S}^{2}/mathbb{S}^{0}$ is homotopy equivalent to $mathbb{S}^{2} vee mathbb{S}^{1}$.. See the link https://www.math3ma.com/blog/clever-homotopy-equivalences?fbclid=IwAR07TYypWvcoz262XQQr2nO3-uG13OU9mO5jhzsQTA6qjfXbYJChxG7EJ88 in Igor Sikora's comment.
To find a space $S$ such that all $S_i$ are deformation retracts of $S$, you can proceed as follows:
Let $f : X to Y$ be a map. Then $Y$ embeds as the base of the mapping cylinder $C(f)$ of $f$ which is a strong deformation retract of $C(f)$. Moreover, $X$ embeds as the top of $C(f)$. It is a well-known that $f$ is a homotopy equivalence if and only if $X$ is a deformation retract of $C(f)$.
Hence any two homotopy equivalent spaces embed as deformation retracts into a bigger space. You can iterate this to finitey many homotopy equivalent spaces.
answered Dec 15 '18 at 10:59
Paul FrostPaul Frost
10.7k3934
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You want to show that all three spaces $S_1, S_2, S_3$ are homotopy equivalent. To do this, it is not necessary to find a space $S$ such that each $S_i$ is a deformation retract (but I shall come back to this point later).
I shall use the following two well-known theorems:
If $A hookrightarrow X$ is a cofibration and $A$ is contractible, then the qotient map $X to X/A$ is a homotopy equivalence.
If $X$ is a CW-complex and $A$ is a subcomplex, then $A hookrightarrow X$ is a cofibration.
Appyling these theorems to $(S_1,diameter)$ and $(S_2,disk)$, we see that both spaces are homotopy equivalent to $S^2/S^0$, where $S^0$ is regarded as the set consisting of north and south pole. That this space is homotopy equivalent to $S_3$ has been shown in Show $mathbb{S}^{2}/mathbb{S}^{0}$ is homotopy equivalent to $mathbb{S}^{2} vee mathbb{S}^{1}$.. See the link https://www.math3ma.com/blog/clever-homotopy-equivalences?fbclid=IwAR07TYypWvcoz262XQQr2nO3-uG13OU9mO5jhzsQTA6qjfXbYJChxG7EJ88 in Igor Sikora's comment.
To find a space $S$ such that all $S_i$ are deformation retracts of $S$, you can proceed as follows:
Let $f : X to Y$ be a map. Then $Y$ embeds as the base of the mapping cylinder $C(f)$ of $f$ which is a strong deformation retract of $C(f)$. Moreover, $X$ embeds as the top of $C(f)$. It is a well-known that $f$ is a homotopy equivalence if and only if $X$ is a deformation retract of $C(f)$.
Hence any two homotopy equivalent spaces embed as deformation retracts into a bigger space. You can iterate this to finitey many homotopy equivalent spaces.
answered Dec 15 '18 at 10:59
Paul FrostPaul Frost
10.7k3934
$endgroup$
add a comment |
$begingroup$
You want to show that all three spaces $S_1, S_2, S_3$ are homotopy equivalent. To do this, it is not necessary to find a space $S$ such that each $S_i$ is a deformation retract (but I shall come back to this point later).
I shall use the following two well-known theorems:
If $A hookrightarrow X$ is a cofibration and $A$ is contractible, then the qotient map $X to X/A$ is a homotopy equivalence.
If $X$ is a CW-complex and $A$ is a subcomplex, then $A hookrightarrow X$ is a cofibration.
Appyling these theorems to $(S_1,diameter)$ and $(S_2,disk)$, we see that both spaces are homotopy equivalent to $S^2/S^0$, where $S^0$ is regarded as the set consisting of north and south pole. That this space is homotopy equivalent to $S_3$ has been shown in Show $mathbb{S}^{2}/mathbb{S}^{0}$ is homotopy equivalent to $mathbb{S}^{2} vee mathbb{S}^{1}$.. See the link https://www.math3ma.com/blog/clever-homotopy-equivalences?fbclid=IwAR07TYypWvcoz262XQQr2nO3-uG13OU9mO5jhzsQTA6qjfXbYJChxG7EJ88 in Igor Sikora's comment.
To find a space $S$ such that all $S_i$ are deformation retracts of $S$, you can proceed as follows:
Let $f : X to Y$ be a map. Then $Y$ embeds as the base of the mapping cylinder $C(f)$ of $f$ which is a strong deformation retract of $C(f)$. Moreover, $X$ embeds as the top of $C(f)$. It is a well-known that $f$ is a homotopy equivalence if and only if $X$ is a deformation retract of $C(f)$.
Hence any two homotopy equivalent spaces embed as deformation retracts into a bigger space. You can iterate this to finitey many homotopy equivalent spaces.
answered Dec 15 '18 at 10:59
Paul FrostPaul Frost
10.7k3934
$endgroup$
add a comment |
$begingroup$
You want to show that all three spaces $S_1, S_2, S_3$ are homotopy equivalent. To do this, it is not necessary to find a space $S$ such that each $S_i$ is a deformation retract (but I shall come back to this point later).
I shall use the following two well-known theorems:
If $A hookrightarrow X$ is a cofibration and $A$ is contractible, then the qotient map $X to X/A$ is a homotopy equivalence.
If $X$ is a CW-complex and $A$ is a subcomplex, then $A hookrightarrow X$ is a cofibration.
Appyling these theorems to $(S_1,diameter)$ and $(S_2,disk)$, we see that both spaces are homotopy equivalent to $S^2/S^0$, where $S^0$ is regarded as the set consisting of north and south pole. That this space is homotopy equivalent to $S_3$ has been shown in Show $mathbb{S}^{2}/mathbb{S}^{0}$ is homotopy equivalent to $mathbb{S}^{2} vee mathbb{S}^{1}$.. See the link https://www.math3ma.com/blog/clever-homotopy-equivalences?fbclid=IwAR07TYypWvcoz262XQQr2nO3-uG13OU9mO5jhzsQTA6qjfXbYJChxG7EJ88 in Igor Sikora's comment.
To find a space $S$ such that all $S_i$ are deformation retracts of $S$, you can proceed as follows:
Let $f : X to Y$ be a map. Then $Y$ embeds as the base of the mapping cylinder $C(f)$ of $f$ which is a strong deformation retract of $C(f)$. Moreover, $X$ embeds as the top of $C(f)$. It is a well-known that $f$ is a homotopy equivalence if and only if $X$ is a deformation retract of $C(f)$.
Hence any two homotopy equivalent spaces embed as deformation retracts into a bigger space. You can iterate this to finitey many homotopy equivalent spaces.
answered Dec 15 '18 at 10:59
Paul FrostPaul Frost
10.7k3934
$endgroup$
You want to show that all three spaces $S_1, S_2, S_3$ are homotopy equivalent. To do this, it is not necessary to find a space $S$ such that each $S_i$ is a deformation retract (but I shall come back to this point later).
I shall use the following two well-known theorems:
If $A hookrightarrow X$ is a cofibration and $A$ is contractible, then the qotient map $X to X/A$ is a homotopy equivalence.
If $X$ is a CW-complex and $A$ is a subcomplex, then $A hookrightarrow X$ is a cofibration.
Appyling these theorems to $(S_1,diameter)$ and $(S_2,disk)$, we see that both spaces are homotopy equivalent to $S^2/S^0$, where $S^0$ is regarded as the set consisting of north and south pole. That this space is homotopy equivalent to $S_3$ has been shown in Show $mathbb{S}^{2}/mathbb{S}^{0}$ is homotopy equivalent to $mathbb{S}^{2} vee mathbb{S}^{1}$.. See the link https://www.math3ma.com/blog/clever-homotopy-equivalences?fbclid=IwAR07TYypWvcoz262XQQr2nO3-uG13OU9mO5jhzsQTA6qjfXbYJChxG7EJ88 in Igor Sikora's comment.
To find a space $S$ such that all $S_i$ are deformation retracts of $S$, you can proceed as follows:
Let $f : X to Y$ be a map. Then $Y$ embeds as the base of the mapping cylinder $C(f)$ of $f$ which is a strong deformation retract of $C(f)$. Moreover, $X$ embeds as the top of $C(f)$. It is a well-known that $f$ is a homotopy equivalence if and only if $X$ is a deformation retract of $C(f)$.
Hence any two homotopy equivalent spaces embed as deformation retracts into a bigger space. You can iterate this to finitey many homotopy equivalent spaces.
answered Dec 15 '18 at 10:59
Paul FrostPaul Frost
10.7k3934
answered Dec 15 '18 at 10:59
Paul FrostPaul Frost
10.7k3934
answered Dec 15 '18 at 10:59
Paul FrostPaul Frost
10.7k3934
answered Dec 15 '18 at 10:59
Paul FrostPaul Frost
10.7k3934
10.7k3934
add a comment |
add a comment |
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$begingroup$
In $(a)$, contract the diameter and squeeze out a disc around the double point to go $(a)Rightarrow (b)$. Also in $(a)$, take one end of the diameter and drag it continuously around the surface of $S^2$ till it coincides with the point at the other end. This gives you $(c)$ with the $S^1$ inside the $S^2$, and this is completely fine.
$endgroup$
– Tyrone
Dec 15 '18 at 10:49
$begingroup$
Similar to math.stackexchange.com/q/3023209.
$endgroup$
– Paul Frost
Dec 15 '18 at 11:00