Prove that $S^n /{p,q}$ is homotopy equivalent to $S^n vee S^1$.
$begingroup$
Prove that $S^n /{p, q}$ homotopy equivalent to $S^n ∨ S^1$
My attempt:
I can see the picture quite clearly: 
But how to write explicit homotopies ?
Thanks in Advance for help!
algebraic-topology homotopy-theory
edited Nov 3 '18 at 15:10
Tyrone
4,50511225
asked Nov 3 '18 at 15:01
CoherentCoherent
1,230523
$endgroup$
add a comment |
$begingroup$
Prove that $S^n /{p, q}$ homotopy equivalent to $S^n ∨ S^1$
My attempt:
I can see the picture quite clearly:
But how to write explicit homotopies ?
Thanks in Advance for help!
algebraic-topology homotopy-theory
edited Nov 3 '18 at 15:10
Tyrone
4,50511225
asked Nov 3 '18 at 15:01
CoherentCoherent
1,230523
$endgroup$
2
$begingroup$
You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
$endgroup$
– Neal
Nov 3 '18 at 15:12
$begingroup$
@Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
$endgroup$
– Cheerful Parsnip
Nov 3 '18 at 16:37
$begingroup$
@Neal Can you kindly post a full answer.
$endgroup$
– Coherent
Nov 5 '18 at 4:40
add a comment |
$begingroup$
Prove that $S^n /{p, q}$ homotopy equivalent to $S^n ∨ S^1$
My attempt:
I can see the picture quite clearly:
But how to write explicit homotopies ?
Thanks in Advance for help!
algebraic-topology homotopy-theory
edited Nov 3 '18 at 15:10
Tyrone
4,50511225
asked Nov 3 '18 at 15:01
CoherentCoherent
1,230523
$endgroup$
Prove that $S^n /{p, q}$ homotopy equivalent to $S^n ∨ S^1$
My attempt:
I can see the picture quite clearly:
But how to write explicit homotopies ?
Thanks in Advance for help!
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
edited Nov 3 '18 at 15:10
Tyrone
4,50511225
asked Nov 3 '18 at 15:01
CoherentCoherent
1,230523
edited Nov 3 '18 at 15:10
Tyrone
4,50511225
asked Nov 3 '18 at 15:01
CoherentCoherent
1,230523
edited Nov 3 '18 at 15:10
Tyrone
4,50511225
edited Nov 3 '18 at 15:10
Tyrone
4,50511225
edited Nov 3 '18 at 15:10
Tyrone
4,50511225
4,50511225
asked Nov 3 '18 at 15:01
CoherentCoherent
1,230523
asked Nov 3 '18 at 15:01
CoherentCoherent
1,230523
asked Nov 3 '18 at 15:01
CoherentCoherent
1,230523
1,230523
2
$begingroup$
You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
$endgroup$
– Neal
Nov 3 '18 at 15:12
$begingroup$
@Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
$endgroup$
– Cheerful Parsnip
Nov 3 '18 at 16:37
$begingroup$
@Neal Can you kindly post a full answer.
$endgroup$
– Coherent
Nov 5 '18 at 4:40
add a comment |
2
$begingroup$
You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
$endgroup$
– Neal
Nov 3 '18 at 15:12
$begingroup$
@Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
$endgroup$
– Cheerful Parsnip
Nov 3 '18 at 16:37
$begingroup$
@Neal Can you kindly post a full answer.
$endgroup$
– Coherent
Nov 5 '18 at 4:40
2
2
$begingroup$
You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
$endgroup$
– Neal
Nov 3 '18 at 15:12
$begingroup$
You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
$endgroup$
– Neal
Nov 3 '18 at 15:12
$begingroup$
@Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
$endgroup$
– Cheerful Parsnip
Nov 3 '18 at 16:37
$begingroup$
@Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
$endgroup$
– Cheerful Parsnip
Nov 3 '18 at 16:37
$begingroup$
@Neal Can you kindly post a full answer.
$endgroup$
– Coherent
Nov 5 '18 at 4:40
$begingroup$
@Neal Can you kindly post a full answer.
$endgroup$
– Coherent
Nov 5 '18 at 4:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $Xsimeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Ysimeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Ysimeq Y/B$, and $Y/B$ is homeomorphic to $S^nvee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $simeq$ you are done.
answered Dec 1 '18 at 7:24
Cheerful ParsnipCheerful Parsnip
21k23397
$endgroup$
add a comment |
$begingroup$
For the inspiration, you should see Example 0.8 in Hatcher's book.
Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 vee S^1$.
You can generalize his argument using $CW$-complex structure of $S^n$.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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oldest
votes
$begingroup$
We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $Xsimeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Ysimeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Ysimeq Y/B$, and $Y/B$ is homeomorphic to $S^nvee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $simeq$ you are done.
answered Dec 1 '18 at 7:24
Cheerful ParsnipCheerful Parsnip
21k23397
$endgroup$
add a comment |
$begingroup$
We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $Xsimeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Ysimeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Ysimeq Y/B$, and $Y/B$ is homeomorphic to $S^nvee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $simeq$ you are done.
answered Dec 1 '18 at 7:24
Cheerful ParsnipCheerful Parsnip
21k23397
$endgroup$
add a comment |
$begingroup$
We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $Xsimeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Ysimeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Ysimeq Y/B$, and $Y/B$ is homeomorphic to $S^nvee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $simeq$ you are done.
answered Dec 1 '18 at 7:24
Cheerful ParsnipCheerful Parsnip
21k23397
$endgroup$
We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $Xsimeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Ysimeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Ysimeq Y/B$, and $Y/B$ is homeomorphic to $S^nvee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $simeq$ you are done.
answered Dec 1 '18 at 7:24
Cheerful ParsnipCheerful Parsnip
21k23397
answered Dec 1 '18 at 7:24
Cheerful ParsnipCheerful Parsnip
21k23397
answered Dec 1 '18 at 7:24
Cheerful ParsnipCheerful Parsnip
21k23397
answered Dec 1 '18 at 7:24
Cheerful ParsnipCheerful Parsnip
21k23397
21k23397
add a comment |
add a comment |
$begingroup$
For the inspiration, you should see Example 0.8 in Hatcher's book.
Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 vee S^1$.
You can generalize his argument using $CW$-complex structure of $S^n$.
$endgroup$
add a comment |
$begingroup$
For the inspiration, you should see Example 0.8 in Hatcher's book.
Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 vee S^1$.
You can generalize his argument using $CW$-complex structure of $S^n$.
$endgroup$
add a comment |
$begingroup$
For the inspiration, you should see Example 0.8 in Hatcher's book.
Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 vee S^1$.
You can generalize his argument using $CW$-complex structure of $S^n$.
$endgroup$
For the inspiration, you should see Example 0.8 in Hatcher's book.
Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 vee S^1$.
You can generalize his argument using $CW$-complex structure of $S^n$.
answered Dec 1 '18 at 8:53
user621469
add a comment |
add a comment |
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2
$begingroup$
You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
$endgroup$
– Neal
Nov 3 '18 at 15:12
$begingroup$
@Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
$endgroup$
– Cheerful Parsnip
Nov 3 '18 at 16:37
$begingroup$
@Neal Can you kindly post a full answer.
$endgroup$
– Coherent
Nov 5 '18 at 4:40