One Point Compactification of $mathbf{R}^n-{0}$
I wanted to know the one point compactification of $mathbf{R}^n-{0}$. The actual problem asks to show that $S^n/{p,q} simeq S^n vee S^1$, where $p,q$ are two distinct points of $S^n$. I know that $S^n-{p,q}$ is homeomorphic to the one point compactification of $mathbf{R}^n-{0}$. So, knowing the result might help to solve the problem. Intuition from $mathbf{R}-{0}$ is not helping much as that is $S^1vee S^1$ and I couldn't work out for even $n=2$. Please help.
general-topology algebraic-topology
asked Nov 24 at 15:33
PSG
3369
add a comment |
I wanted to know the one point compactification of $mathbf{R}^n-{0}$. The actual problem asks to show that $S^n/{p,q} simeq S^n vee S^1$, where $p,q$ are two distinct points of $S^n$. I know that $S^n-{p,q}$ is homeomorphic to the one point compactification of $mathbf{R}^n-{0}$. So, knowing the result might help to solve the problem. Intuition from $mathbf{R}-{0}$ is not helping much as that is $S^1vee S^1$ and I couldn't work out for even $n=2$. Please help.
general-topology algebraic-topology
asked Nov 24 at 15:33
PSG
3369
In general, the way to make a one-point compactification is to identify the limits of all sequences that you really want to converge. In this case, it would be adding a point at infinity and identifying it with $0$. If you think about the sphere $S^n$ and stereographic projection, this would correspond to identifying the north and south poles.
– Michael Burr
Nov 24 at 15:36
Ok, is it a known space? My guess would be something involved with wedge product.
– PSG
Nov 24 at 15:39
Perhaps the example on page 11 on Hatcher's Algebraic Topology would be helpful.
– Michael Burr
Nov 24 at 15:45
Sorry, I am not really comfortable with CW-Cells. Is there a solution with precise maps?
– PSG
Nov 24 at 15:52
add a comment |
I wanted to know the one point compactification of $mathbf{R}^n-{0}$. The actual problem asks to show that $S^n/{p,q} simeq S^n vee S^1$, where $p,q$ are two distinct points of $S^n$. I know that $S^n-{p,q}$ is homeomorphic to the one point compactification of $mathbf{R}^n-{0}$. So, knowing the result might help to solve the problem. Intuition from $mathbf{R}-{0}$ is not helping much as that is $S^1vee S^1$ and I couldn't work out for even $n=2$. Please help.
general-topology algebraic-topology
asked Nov 24 at 15:33
PSG
3369
I wanted to know the one point compactification of $mathbf{R}^n-{0}$. The actual problem asks to show that $S^n/{p,q} simeq S^n vee S^1$, where $p,q$ are two distinct points of $S^n$. I know that $S^n-{p,q}$ is homeomorphic to the one point compactification of $mathbf{R}^n-{0}$. So, knowing the result might help to solve the problem. Intuition from $mathbf{R}-{0}$ is not helping much as that is $S^1vee S^1$ and I couldn't work out for even $n=2$. Please help.
general-topology algebraic-topology
general-topology algebraic-topology
asked Nov 24 at 15:33
PSG
3369
asked Nov 24 at 15:33
PSG
3369
asked Nov 24 at 15:33
PSG
3369
asked Nov 24 at 15:33
PSG
3369
asked Nov 24 at 15:33
PSG
3369
3369
In general, the way to make a one-point compactification is to identify the limits of all sequences that you really want to converge. In this case, it would be adding a point at infinity and identifying it with $0$. If you think about the sphere $S^n$ and stereographic projection, this would correspond to identifying the north and south poles.
– Michael Burr
Nov 24 at 15:36
Ok, is it a known space? My guess would be something involved with wedge product.
– PSG
Nov 24 at 15:39
Perhaps the example on page 11 on Hatcher's Algebraic Topology would be helpful.
– Michael Burr
Nov 24 at 15:45
Sorry, I am not really comfortable with CW-Cells. Is there a solution with precise maps?
– PSG
Nov 24 at 15:52
add a comment |
In general, the way to make a one-point compactification is to identify the limits of all sequences that you really want to converge. In this case, it would be adding a point at infinity and identifying it with $0$. If you think about the sphere $S^n$ and stereographic projection, this would correspond to identifying the north and south poles.
– Michael Burr
Nov 24 at 15:36
Ok, is it a known space? My guess would be something involved with wedge product.
– PSG
Nov 24 at 15:39
Perhaps the example on page 11 on Hatcher's Algebraic Topology would be helpful.
– Michael Burr
Nov 24 at 15:45
Sorry, I am not really comfortable with CW-Cells. Is there a solution with precise maps?
– PSG
Nov 24 at 15:52
In general, the way to make a one-point compactification is to identify the limits of all sequences that you really want to converge. In this case, it would be adding a point at infinity and identifying it with $0$. If you think about the sphere $S^n$ and stereographic projection, this would correspond to identifying the north and south poles.
– Michael Burr
Nov 24 at 15:36
In general, the way to make a one-point compactification is to identify the limits of all sequences that you really want to converge. In this case, it would be adding a point at infinity and identifying it with $0$. If you think about the sphere $S^n$ and stereographic projection, this would correspond to identifying the north and south poles.
– Michael Burr
Nov 24 at 15:36
Ok, is it a known space? My guess would be something involved with wedge product.
– PSG
Nov 24 at 15:39
Ok, is it a known space? My guess would be something involved with wedge product.
– PSG
Nov 24 at 15:39
Perhaps the example on page 11 on Hatcher's Algebraic Topology would be helpful.
– Michael Burr
Nov 24 at 15:45
Perhaps the example on page 11 on Hatcher's Algebraic Topology would be helpful.
– Michael Burr
Nov 24 at 15:45
Sorry, I am not really comfortable with CW-Cells. Is there a solution with precise maps?
– PSG
Nov 24 at 15:52
Sorry, I am not really comfortable with CW-Cells. Is there a solution with precise maps?
– PSG
Nov 24 at 15:52
add a comment |
1 Answer
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By stereographic projection, you can turn $Bbb R^n-{0}$ into $S^n-{N,S}$. You want to make this compact by adding a single point. We wish to add $N$ to fix problems near the north pole and $S$ for problems near the south pole. So to solve both problems at once, we "pinch" $N$ and $S$ together and then only add a single point...
answered Nov 24 at 15:39
Hagen von Eitzen
275k21268495
I got the idea, so what would be the final space?
– PSG
Nov 24 at 15:41
1
@PSG A sphere with the north and south poles identified. This is homotopy equivalent to the sphere with a circle attached at a point.
– Aleksandar Milivojevic
Nov 24 at 15:45
For $S^1$ , I can see that if we pinch the $N$ and $S$ together we get $S^1vee S^1$, it's difficult to see in higher dimension , maybe you can give a closed form of the space. Is it $S^nvee S^1$?
– PSG
Nov 24 at 15:46
1
Draw a line from N to S through the “hole” in the sphere. (Contracting the line gives you pinching, but now we want the line to be there.) Now move N along the sphere over to S, and see that the line becomes a circle.
– Aleksandar Milivojevic
Nov 24 at 15:58
add a comment |
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By stereographic projection, you can turn $Bbb R^n-{0}$ into $S^n-{N,S}$. You want to make this compact by adding a single point. We wish to add $N$ to fix problems near the north pole and $S$ for problems near the south pole. So to solve both problems at once, we "pinch" $N$ and $S$ together and then only add a single point...
answered Nov 24 at 15:39
Hagen von Eitzen
275k21268495
I got the idea, so what would be the final space?
– PSG
Nov 24 at 15:41
1
@PSG A sphere with the north and south poles identified. This is homotopy equivalent to the sphere with a circle attached at a point.
– Aleksandar Milivojevic
Nov 24 at 15:45
For $S^1$ , I can see that if we pinch the $N$ and $S$ together we get $S^1vee S^1$, it's difficult to see in higher dimension , maybe you can give a closed form of the space. Is it $S^nvee S^1$?
– PSG
Nov 24 at 15:46
1
Draw a line from N to S through the “hole” in the sphere. (Contracting the line gives you pinching, but now we want the line to be there.) Now move N along the sphere over to S, and see that the line becomes a circle.
– Aleksandar Milivojevic
Nov 24 at 15:58
add a comment |
By stereographic projection, you can turn $Bbb R^n-{0}$ into $S^n-{N,S}$. You want to make this compact by adding a single point. We wish to add $N$ to fix problems near the north pole and $S$ for problems near the south pole. So to solve both problems at once, we "pinch" $N$ and $S$ together and then only add a single point...
answered Nov 24 at 15:39
Hagen von Eitzen
275k21268495
I got the idea, so what would be the final space?
– PSG
Nov 24 at 15:41
1
@PSG A sphere with the north and south poles identified. This is homotopy equivalent to the sphere with a circle attached at a point.
– Aleksandar Milivojevic
Nov 24 at 15:45
For $S^1$ , I can see that if we pinch the $N$ and $S$ together we get $S^1vee S^1$, it's difficult to see in higher dimension , maybe you can give a closed form of the space. Is it $S^nvee S^1$?
– PSG
Nov 24 at 15:46
1
Draw a line from N to S through the “hole” in the sphere. (Contracting the line gives you pinching, but now we want the line to be there.) Now move N along the sphere over to S, and see that the line becomes a circle.
– Aleksandar Milivojevic
Nov 24 at 15:58
add a comment |
By stereographic projection, you can turn $Bbb R^n-{0}$ into $S^n-{N,S}$. You want to make this compact by adding a single point. We wish to add $N$ to fix problems near the north pole and $S$ for problems near the south pole. So to solve both problems at once, we "pinch" $N$ and $S$ together and then only add a single point...
answered Nov 24 at 15:39
Hagen von Eitzen
275k21268495
By stereographic projection, you can turn $Bbb R^n-{0}$ into $S^n-{N,S}$. You want to make this compact by adding a single point. We wish to add $N$ to fix problems near the north pole and $S$ for problems near the south pole. So to solve both problems at once, we "pinch" $N$ and $S$ together and then only add a single point...
answered Nov 24 at 15:39
Hagen von Eitzen
275k21268495
answered Nov 24 at 15:39
Hagen von Eitzen
275k21268495
answered Nov 24 at 15:39
Hagen von Eitzen
275k21268495
answered Nov 24 at 15:39
Hagen von Eitzen
275k21268495
275k21268495
I got the idea, so what would be the final space?
– PSG
Nov 24 at 15:41
1
@PSG A sphere with the north and south poles identified. This is homotopy equivalent to the sphere with a circle attached at a point.
– Aleksandar Milivojevic
Nov 24 at 15:45
For $S^1$ , I can see that if we pinch the $N$ and $S$ together we get $S^1vee S^1$, it's difficult to see in higher dimension , maybe you can give a closed form of the space. Is it $S^nvee S^1$?
– PSG
Nov 24 at 15:46
1
Draw a line from N to S through the “hole” in the sphere. (Contracting the line gives you pinching, but now we want the line to be there.) Now move N along the sphere over to S, and see that the line becomes a circle.
– Aleksandar Milivojevic
Nov 24 at 15:58
add a comment |
I got the idea, so what would be the final space?
– PSG
Nov 24 at 15:41
1
@PSG A sphere with the north and south poles identified. This is homotopy equivalent to the sphere with a circle attached at a point.
– Aleksandar Milivojevic
Nov 24 at 15:45
For $S^1$ , I can see that if we pinch the $N$ and $S$ together we get $S^1vee S^1$, it's difficult to see in higher dimension , maybe you can give a closed form of the space. Is it $S^nvee S^1$?
– PSG
Nov 24 at 15:46
1
Draw a line from N to S through the “hole” in the sphere. (Contracting the line gives you pinching, but now we want the line to be there.) Now move N along the sphere over to S, and see that the line becomes a circle.
– Aleksandar Milivojevic
Nov 24 at 15:58
I got the idea, so what would be the final space?
– PSG
Nov 24 at 15:41
I got the idea, so what would be the final space?
– PSG
Nov 24 at 15:41
1
1
@PSG A sphere with the north and south poles identified. This is homotopy equivalent to the sphere with a circle attached at a point.
– Aleksandar Milivojevic
Nov 24 at 15:45
@PSG A sphere with the north and south poles identified. This is homotopy equivalent to the sphere with a circle attached at a point.
– Aleksandar Milivojevic
Nov 24 at 15:45
For $S^1$ , I can see that if we pinch the $N$ and $S$ together we get $S^1vee S^1$, it's difficult to see in higher dimension , maybe you can give a closed form of the space. Is it $S^nvee S^1$?
– PSG
Nov 24 at 15:46
For $S^1$ , I can see that if we pinch the $N$ and $S$ together we get $S^1vee S^1$, it's difficult to see in higher dimension , maybe you can give a closed form of the space. Is it $S^nvee S^1$?
– PSG
Nov 24 at 15:46
1
1
Draw a line from N to S through the “hole” in the sphere. (Contracting the line gives you pinching, but now we want the line to be there.) Now move N along the sphere over to S, and see that the line becomes a circle.
– Aleksandar Milivojevic
Nov 24 at 15:58
Draw a line from N to S through the “hole” in the sphere. (Contracting the line gives you pinching, but now we want the line to be there.) Now move N along the sphere over to S, and see that the line becomes a circle.
– Aleksandar Milivojevic
Nov 24 at 15:58
add a comment |
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In general, the way to make a one-point compactification is to identify the limits of all sequences that you really want to converge. In this case, it would be adding a point at infinity and identifying it with $0$. If you think about the sphere $S^n$ and stereographic projection, this would correspond to identifying the north and south poles.
– Michael Burr
Nov 24 at 15:36
Ok, is it a known space? My guess would be something involved with wedge product.
– PSG
Nov 24 at 15:39
Perhaps the example on page 11 on Hatcher's Algebraic Topology would be helpful.
– Michael Burr
Nov 24 at 15:45
Sorry, I am not really comfortable with CW-Cells. Is there a solution with precise maps?
– PSG
Nov 24 at 15:52