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Let $X=mathbbR^2-0$ and $Y=S^1 cup[0,1]times0$. Prove or Disprove that $X$ is homeomorphic to $Y$.
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up vote
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Let $X=mathbbR^2-0$ and $Y=S^1 cup[0,1]times0$.
Prove or Disprove that $X$ is homeomorphic to $Y$.
My thoughts:
Since X is the punctured plane, in terms of retraction, I believe that $S^1$ is a deformation retact of $X$, Similarly, the space $Y$ is the unit circle with an extra "strip" along the $x- axis$ same as the radius. And so $S^1$ is a deformation retract of $Y$ as well.
As a result, the fundamental groups of both $X$ and $Y$ are isomorphic to the fundamental group of the unit circle $pi_1(X,x_0) approx mathbbZ approx pi_1(Y,y_0)$ .
Does this isomorphism guarantee a homeomorphism between the spaces $X$ and $Y$?
if not, Is there a way to appraoch it without involving fundamental groups?
general-topology algebraic-topology
asked Aug 6 at 2:49
J. Kyei
1758
add a comment |Â
up vote
5
down vote
favorite
Let $X=mathbbR^2-0$ and $Y=S^1 cup[0,1]times0$.
Prove or Disprove that $X$ is homeomorphic to $Y$.
My thoughts:
Since X is the punctured plane, in terms of retraction, I believe that $S^1$ is a deformation retact of $X$, Similarly, the space $Y$ is the unit circle with an extra "strip" along the $x- axis$ same as the radius. And so $S^1$ is a deformation retract of $Y$ as well.
As a result, the fundamental groups of both $X$ and $Y$ are isomorphic to the fundamental group of the unit circle $pi_1(X,x_0) approx mathbbZ approx pi_1(Y,y_0)$ .
Does this isomorphism guarantee a homeomorphism between the spaces $X$ and $Y$?
if not, Is there a way to appraoch it without involving fundamental groups?
general-topology algebraic-topology
asked Aug 6 at 2:49
J. Kyei
1758
With respect to your last question: non-homeomorphic spaces may have isomorphic fundamental groups, like a point and the unit disk, for example (or your two spaces!)
– Pedro Tamaroff♦
Aug 6 at 3:22
Also with respect to your last question: A famous problem of mathematics is: Suppose $M$ is a three-dimensional manifold with no boundary and a trivial fundamental group. Must $M$ be homeomorphic to $S^3$? (A three-dimensional manifold is a space that is everywhere locally homeomorphic to $Bbb R^3$.) This question was unanswered for nearly a hundred years! (See: Poincaré conjecture.) The corresponding question about $S^2$ was resolved long ago, but for $S^4$ the answer is still unknown!
– MJD
Aug 6 at 19:07
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $X=mathbbR^2-0$ and $Y=S^1 cup[0,1]times0$.
Prove or Disprove that $X$ is homeomorphic to $Y$.
My thoughts:
Since X is the punctured plane, in terms of retraction, I believe that $S^1$ is a deformation retact of $X$, Similarly, the space $Y$ is the unit circle with an extra "strip" along the $x- axis$ same as the radius. And so $S^1$ is a deformation retract of $Y$ as well.
As a result, the fundamental groups of both $X$ and $Y$ are isomorphic to the fundamental group of the unit circle $pi_1(X,x_0) approx mathbbZ approx pi_1(Y,y_0)$ .
Does this isomorphism guarantee a homeomorphism between the spaces $X$ and $Y$?
if not, Is there a way to appraoch it without involving fundamental groups?
general-topology algebraic-topology
asked Aug 6 at 2:49
J. Kyei
1758
Let $X=mathbbR^2-0$ and $Y=S^1 cup[0,1]times0$.
Prove or Disprove that $X$ is homeomorphic to $Y$.
My thoughts:
Since X is the punctured plane, in terms of retraction, I believe that $S^1$ is a deformation retact of $X$, Similarly, the space $Y$ is the unit circle with an extra "strip" along the $x- axis$ same as the radius. And so $S^1$ is a deformation retract of $Y$ as well.
As a result, the fundamental groups of both $X$ and $Y$ are isomorphic to the fundamental group of the unit circle $pi_1(X,x_0) approx mathbbZ approx pi_1(Y,y_0)$ .
Does this isomorphism guarantee a homeomorphism between the spaces $X$ and $Y$?
if not, Is there a way to appraoch it without involving fundamental groups?
general-topology algebraic-topology
asked Aug 6 at 2:49
J. Kyei
1758
asked Aug 6 at 2:49
J. Kyei
1758
asked Aug 6 at 2:49
J. Kyei
1758
asked Aug 6 at 2:49
J. Kyei
1758
1758
With respect to your last question: non-homeomorphic spaces may have isomorphic fundamental groups, like a point and the unit disk, for example (or your two spaces!)
– Pedro Tamaroff♦
Aug 6 at 3:22
Also with respect to your last question: A famous problem of mathematics is: Suppose $M$ is a three-dimensional manifold with no boundary and a trivial fundamental group. Must $M$ be homeomorphic to $S^3$? (A three-dimensional manifold is a space that is everywhere locally homeomorphic to $Bbb R^3$.) This question was unanswered for nearly a hundred years! (See: Poincaré conjecture.) The corresponding question about $S^2$ was resolved long ago, but for $S^4$ the answer is still unknown!
– MJD
Aug 6 at 19:07
add a comment |Â
With respect to your last question: non-homeomorphic spaces may have isomorphic fundamental groups, like a point and the unit disk, for example (or your two spaces!)
– Pedro Tamaroff♦
Aug 6 at 3:22
Also with respect to your last question: A famous problem of mathematics is: Suppose $M$ is a three-dimensional manifold with no boundary and a trivial fundamental group. Must $M$ be homeomorphic to $S^3$? (A three-dimensional manifold is a space that is everywhere locally homeomorphic to $Bbb R^3$.) This question was unanswered for nearly a hundred years! (See: Poincaré conjecture.) The corresponding question about $S^2$ was resolved long ago, but for $S^4$ the answer is still unknown!
– MJD
Aug 6 at 19:07
With respect to your last question: non-homeomorphic spaces may have isomorphic fundamental groups, like a point and the unit disk, for example (or your two spaces!)
– Pedro Tamaroff♦
Aug 6 at 3:22
With respect to your last question: non-homeomorphic spaces may have isomorphic fundamental groups, like a point and the unit disk, for example (or your two spaces!)
– Pedro Tamaroff♦
Aug 6 at 3:22
Also with respect to your last question: A famous problem of mathematics is: Suppose $M$ is a three-dimensional manifold with no boundary and a trivial fundamental group. Must $M$ be homeomorphic to $S^3$? (A three-dimensional manifold is a space that is everywhere locally homeomorphic to $Bbb R^3$.) This question was unanswered for nearly a hundred years! (See: Poincaré conjecture.) The corresponding question about $S^2$ was resolved long ago, but for $S^4$ the answer is still unknown!
– MJD
Aug 6 at 19:07
Also with respect to your last question: A famous problem of mathematics is: Suppose $M$ is a three-dimensional manifold with no boundary and a trivial fundamental group. Must $M$ be homeomorphic to $S^3$? (A three-dimensional manifold is a space that is everywhere locally homeomorphic to $Bbb R^3$.) This question was unanswered for nearly a hundred years! (See: Poincaré conjecture.) The corresponding question about $S^2$ was resolved long ago, but for $S^4$ the answer is still unknown!
– MJD
Aug 6 at 19:07
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
You can disconnect $Y$ by removing two points but if you remove two points to $X$ it is always connected.
answered Aug 6 at 2:51
Tsemo Aristide
51.4k11243
And since Taking two points from X does not make it disconnected, they cannot be homeomorphic?
– J. Kyei
Aug 6 at 2:53
yes, they cannot. If $f:Xrightarrow Y$ is an homeomorphism, its restriction to $X-x_1,...,x_prightarrow Y-f(x_1),...,f(x_p)$ is also an homeomorphism.
– Tsemo Aristide
Aug 6 at 2:56
Thank you Sir! I I understand now
– J. Kyei
Aug 6 at 3:00
1
Why two points ? I thought we could remove $(1,0)$ from $Y$ to make it disconnected.
– Sou
Aug 6 at 6:48
add a comment |Â
up vote
5
down vote
$Y$ is compact, but $X$ isn't.
But $X$ and $Y$ are homotopy equivalent. Both have $S^1$ as
a deformation retract.
answered Aug 6 at 2:54
Lord Shark the Unknown
86k951112
1
I realized. I guess this shows isomorphism between fundamental group does not necessarily say the space are homeomorphic. Thank you
– J. Kyei
Aug 6 at 3:01
add a comment |Â
up vote
2
down vote
$HINT:$ Deleting the point $(1,0)$ from $Y$ results it disconnected. Can you find such point in $X$?
answered Aug 6 at 4:02
Mathlover
3,5831021
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
You can disconnect $Y$ by removing two points but if you remove two points to $X$ it is always connected.
answered Aug 6 at 2:51
Tsemo Aristide
51.4k11243
And since Taking two points from X does not make it disconnected, they cannot be homeomorphic?
– J. Kyei
Aug 6 at 2:53
yes, they cannot. If $f:Xrightarrow Y$ is an homeomorphism, its restriction to $X-x_1,...,x_prightarrow Y-f(x_1),...,f(x_p)$ is also an homeomorphism.
– Tsemo Aristide
Aug 6 at 2:56
Thank you Sir! I I understand now
– J. Kyei
Aug 6 at 3:00
1
Why two points ? I thought we could remove $(1,0)$ from $Y$ to make it disconnected.
– Sou
Aug 6 at 6:48
add a comment |Â
up vote
5
down vote
You can disconnect $Y$ by removing two points but if you remove two points to $X$ it is always connected.
answered Aug 6 at 2:51
Tsemo Aristide
51.4k11243
And since Taking two points from X does not make it disconnected, they cannot be homeomorphic?
– J. Kyei
Aug 6 at 2:53
yes, they cannot. If $f:Xrightarrow Y$ is an homeomorphism, its restriction to $X-x_1,...,x_prightarrow Y-f(x_1),...,f(x_p)$ is also an homeomorphism.
– Tsemo Aristide
Aug 6 at 2:56
Thank you Sir! I I understand now
– J. Kyei
Aug 6 at 3:00
1
Why two points ? I thought we could remove $(1,0)$ from $Y$ to make it disconnected.
– Sou
Aug 6 at 6:48
add a comment |Â
up vote
5
down vote
up vote
5
down vote
You can disconnect $Y$ by removing two points but if you remove two points to $X$ it is always connected.
answered Aug 6 at 2:51
Tsemo Aristide
51.4k11243
You can disconnect $Y$ by removing two points but if you remove two points to $X$ it is always connected.
answered Aug 6 at 2:51
Tsemo Aristide
51.4k11243
answered Aug 6 at 2:51
Tsemo Aristide
51.4k11243
answered Aug 6 at 2:51
Tsemo Aristide
51.4k11243
answered Aug 6 at 2:51
Tsemo Aristide
51.4k11243
51.4k11243
And since Taking two points from X does not make it disconnected, they cannot be homeomorphic?
– J. Kyei
Aug 6 at 2:53
yes, they cannot. If $f:Xrightarrow Y$ is an homeomorphism, its restriction to $X-x_1,...,x_prightarrow Y-f(x_1),...,f(x_p)$ is also an homeomorphism.
– Tsemo Aristide
Aug 6 at 2:56
Thank you Sir! I I understand now
– J. Kyei
Aug 6 at 3:00
1
Why two points ? I thought we could remove $(1,0)$ from $Y$ to make it disconnected.
– Sou
Aug 6 at 6:48
add a comment |Â
And since Taking two points from X does not make it disconnected, they cannot be homeomorphic?
– J. Kyei
Aug 6 at 2:53
yes, they cannot. If $f:Xrightarrow Y$ is an homeomorphism, its restriction to $X-x_1,...,x_prightarrow Y-f(x_1),...,f(x_p)$ is also an homeomorphism.
– Tsemo Aristide
Aug 6 at 2:56
Thank you Sir! I I understand now
– J. Kyei
Aug 6 at 3:00
1
Why two points ? I thought we could remove $(1,0)$ from $Y$ to make it disconnected.
– Sou
Aug 6 at 6:48
And since Taking two points from X does not make it disconnected, they cannot be homeomorphic?
– J. Kyei
Aug 6 at 2:53
And since Taking two points from X does not make it disconnected, they cannot be homeomorphic?
– J. Kyei
Aug 6 at 2:53
yes, they cannot. If $f:Xrightarrow Y$ is an homeomorphism, its restriction to $X-x_1,...,x_prightarrow Y-f(x_1),...,f(x_p)$ is also an homeomorphism.
– Tsemo Aristide
Aug 6 at 2:56
yes, they cannot. If $f:Xrightarrow Y$ is an homeomorphism, its restriction to $X-x_1,...,x_prightarrow Y-f(x_1),...,f(x_p)$ is also an homeomorphism.
– Tsemo Aristide
Aug 6 at 2:56
Thank you Sir! I I understand now
– J. Kyei
Aug 6 at 3:00
Thank you Sir! I I understand now
– J. Kyei
Aug 6 at 3:00
1
1
Why two points ? I thought we could remove $(1,0)$ from $Y$ to make it disconnected.
– Sou
Aug 6 at 6:48
Why two points ? I thought we could remove $(1,0)$ from $Y$ to make it disconnected.
– Sou
Aug 6 at 6:48
add a comment |Â
up vote
5
down vote
$Y$ is compact, but $X$ isn't.
But $X$ and $Y$ are homotopy equivalent. Both have $S^1$ as
a deformation retract.
answered Aug 6 at 2:54
Lord Shark the Unknown
86k951112
1
I realized. I guess this shows isomorphism between fundamental group does not necessarily say the space are homeomorphic. Thank you
– J. Kyei
Aug 6 at 3:01
add a comment |Â
up vote
5
down vote
$Y$ is compact, but $X$ isn't.
But $X$ and $Y$ are homotopy equivalent. Both have $S^1$ as
a deformation retract.
answered Aug 6 at 2:54
Lord Shark the Unknown
86k951112
1
I realized. I guess this shows isomorphism between fundamental group does not necessarily say the space are homeomorphic. Thank you
– J. Kyei
Aug 6 at 3:01
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$Y$ is compact, but $X$ isn't.
But $X$ and $Y$ are homotopy equivalent. Both have $S^1$ as
a deformation retract.
answered Aug 6 at 2:54
Lord Shark the Unknown
86k951112
$Y$ is compact, but $X$ isn't.
But $X$ and $Y$ are homotopy equivalent. Both have $S^1$ as
a deformation retract.
answered Aug 6 at 2:54
Lord Shark the Unknown
86k951112
answered Aug 6 at 2:54
Lord Shark the Unknown
86k951112
answered Aug 6 at 2:54
Lord Shark the Unknown
86k951112
answered Aug 6 at 2:54
Lord Shark the Unknown
86k951112
86k951112
1
I realized. I guess this shows isomorphism between fundamental group does not necessarily say the space are homeomorphic. Thank you
– J. Kyei
Aug 6 at 3:01
add a comment |Â
1
I realized. I guess this shows isomorphism between fundamental group does not necessarily say the space are homeomorphic. Thank you
– J. Kyei
Aug 6 at 3:01
1
1
I realized. I guess this shows isomorphism between fundamental group does not necessarily say the space are homeomorphic. Thank you
– J. Kyei
Aug 6 at 3:01
I realized. I guess this shows isomorphism between fundamental group does not necessarily say the space are homeomorphic. Thank you
– J. Kyei
Aug 6 at 3:01
add a comment |Â
up vote
2
down vote
$HINT:$ Deleting the point $(1,0)$ from $Y$ results it disconnected. Can you find such point in $X$?
answered Aug 6 at 4:02
Mathlover
3,5831021
add a comment |Â
up vote
2
down vote
$HINT:$ Deleting the point $(1,0)$ from $Y$ results it disconnected. Can you find such point in $X$?
answered Aug 6 at 4:02
Mathlover
3,5831021
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$HINT:$ Deleting the point $(1,0)$ from $Y$ results it disconnected. Can you find such point in $X$?
answered Aug 6 at 4:02
Mathlover
3,5831021
$HINT:$ Deleting the point $(1,0)$ from $Y$ results it disconnected. Can you find such point in $X$?
answered Aug 6 at 4:02
Mathlover
3,5831021
answered Aug 6 at 4:02
Mathlover
3,5831021
answered Aug 6 at 4:02
Mathlover
3,5831021
answered Aug 6 at 4:02
Mathlover
3,5831021
3,5831021
add a comment |Â
add a comment |Â
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With respect to your last question: non-homeomorphic spaces may have isomorphic fundamental groups, like a point and the unit disk, for example (or your two spaces!)
– Pedro Tamaroff♦
Aug 6 at 3:22
Also with respect to your last question: A famous problem of mathematics is: Suppose $M$ is a three-dimensional manifold with no boundary and a trivial fundamental group. Must $M$ be homeomorphic to $S^3$? (A three-dimensional manifold is a space that is everywhere locally homeomorphic to $Bbb R^3$.) This question was unanswered for nearly a hundred years! (See: Poincaré conjecture.) The corresponding question about $S^2$ was resolved long ago, but for $S^4$ the answer is still unknown!
– MJD
Aug 6 at 19:07