Mixing
If the connected sum $A#B$ is homeomorphic to $S^2$ then $Acong B cong S^2$
$begingroup$
I was looking for this, but I can't find anything.
Problem. Let $A, B$ two compact surfaces such that $A#B cong S^2$ then $Acong B cong S^2$.
I considerd infinite connected sum $A#B#Adotsc$ and $B#A#Bdotsc$ These are homeomorphic to $mathbb{R}^2$ and with ${infty}$ to $S^2$, but I cant finalize the prove.
Can you get me a hint? Thank
Edit. Can I prove it using this? Connected sum of non orientable surfaces is non orienable, then A and B are orientables, in fact, are respectively connected sum of n and m torus. Hence we also know that connected sum of A and B is homeomorphic to the connected sum of $n+m$ torus. Therefore $m+n=0$ and $m,ngeq 0$ so $m=n=0$.
general-topology algebraic-topology surfaces
edited May 3 '17 at 13:05
Adam Hughes
32.2k83770
asked May 3 '17 at 11:56
Rafael Gonzalez LopezRafael Gonzalez Lopez
677212
$endgroup$
add a comment |
$begingroup$
I was looking for this, but I can't find anything.
Problem. Let $A, B$ two compact surfaces such that $A#B cong S^2$ then $Acong B cong S^2$.
I considerd infinite connected sum $A#B#Adotsc$ and $B#A#Bdotsc$ These are homeomorphic to $mathbb{R}^2$ and with ${infty}$ to $S^2$, but I cant finalize the prove.
Can you get me a hint? Thank
Edit. Can I prove it using this? Connected sum of non orientable surfaces is non orienable, then A and B are orientables, in fact, are respectively connected sum of n and m torus. Hence we also know that connected sum of A and B is homeomorphic to the connected sum of $n+m$ torus. Therefore $m+n=0$ and $m,ngeq 0$ so $m=n=0$.
general-topology algebraic-topology surfaces
edited May 3 '17 at 13:05
Adam Hughes
32.2k83770
asked May 3 '17 at 11:56
Rafael Gonzalez LopezRafael Gonzalez Lopez
677212
$endgroup$
add a comment |
$begingroup$
I was looking for this, but I can't find anything.
Problem. Let $A, B$ two compact surfaces such that $A#B cong S^2$ then $Acong B cong S^2$.
I considerd infinite connected sum $A#B#Adotsc$ and $B#A#Bdotsc$ These are homeomorphic to $mathbb{R}^2$ and with ${infty}$ to $S^2$, but I cant finalize the prove.
Can you get me a hint? Thank
Edit. Can I prove it using this? Connected sum of non orientable surfaces is non orienable, then A and B are orientables, in fact, are respectively connected sum of n and m torus. Hence we also know that connected sum of A and B is homeomorphic to the connected sum of $n+m$ torus. Therefore $m+n=0$ and $m,ngeq 0$ so $m=n=0$.
general-topology algebraic-topology surfaces
edited May 3 '17 at 13:05
Adam Hughes
32.2k83770
asked May 3 '17 at 11:56
Rafael Gonzalez LopezRafael Gonzalez Lopez
677212
$endgroup$
I was looking for this, but I can't find anything.
Problem. Let $A, B$ two compact surfaces such that $A#B cong S^2$ then $Acong B cong S^2$.
I considerd infinite connected sum $A#B#Adotsc$ and $B#A#Bdotsc$ These are homeomorphic to $mathbb{R}^2$ and with ${infty}$ to $S^2$, but I cant finalize the prove.
Can you get me a hint? Thank
Edit. Can I prove it using this? Connected sum of non orientable surfaces is non orienable, then A and B are orientables, in fact, are respectively connected sum of n and m torus. Hence we also know that connected sum of A and B is homeomorphic to the connected sum of $n+m$ torus. Therefore $m+n=0$ and $m,ngeq 0$ so $m=n=0$.
general-topology algebraic-topology surfaces
general-topology algebraic-topology surfaces
edited May 3 '17 at 13:05
Adam Hughes
32.2k83770
asked May 3 '17 at 11:56
Rafael Gonzalez LopezRafael Gonzalez Lopez
677212
edited May 3 '17 at 13:05
Adam Hughes
32.2k83770
asked May 3 '17 at 11:56
Rafael Gonzalez LopezRafael Gonzalez Lopez
677212
edited May 3 '17 at 13:05
Adam Hughes
32.2k83770
edited May 3 '17 at 13:05
Adam Hughes
32.2k83770
edited May 3 '17 at 13:05
Adam Hughes
32.2k83770
32.2k83770
asked May 3 '17 at 11:56
Rafael Gonzalez LopezRafael Gonzalez Lopez
677212
asked May 3 '17 at 11:56
Rafael Gonzalez LopezRafael Gonzalez Lopez
677212
asked May 3 '17 at 11:56
Rafael Gonzalez LopezRafael Gonzalez Lopez
677212
677212
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Compact surfaces form a (commutative) monoid under connected sum where $S^2$ is the identity. If you don't already know that, just consider what connected sum does, it cuts out a disk and identifies the boundary, so if you have anything non-orientable it stays that way, and if you have any handles, it only increases the number of handles.
answered May 3 '17 at 11:59
Adam HughesAdam Hughes
32.2k83770
$endgroup$
$begingroup$
Can you see my edit? Thank you.
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:20
$begingroup$
@RafaelGonzalezLopez as far as your edit goes: yes you can. I didn't appeal to the classification theorem because you didn't indicate whether or not you had already learned it, but with that it is easy as you say.
$endgroup$
– Adam Hughes
May 3 '17 at 13:05
add a comment |
$begingroup$
As you have mentioned, $A#(B#A#B#cdots)simeq mathbb{R}^2$ and $B#A#B#cdotssimeq mathbb{R}^2$.
But observe that if $X$ is a compact surface, $X#mathbb{R}^2 simeq Xsetminus{x_0}$ (=$X$ minus a point) Therefore, it means that $A$ minus a point is homeomorphic to $mathbb{R}^2$, and then it is easy to conclude that $A simeq S^2$. That $Bsimeq S^2$ follows easily by symmetry.
answered May 3 '17 at 12:06
Henry ParkHenry Park
1,710724
$endgroup$
$begingroup$
That was the point! Thank you for comment, but I would rather prove it without using inifinite sum ^^
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:27
1
$begingroup$
@RafaelGonzalezLopez In fact, it is a very famous argument due to Barry Mazur. Exactly the same argument can be applied to prove some other theorems, such as "the only invertible elements in the monoid of knots and their knot sums is the unknot". See, for instance, math.stonybrook.edu/~jack/MilnorOslo-print.pdf or en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle
$endgroup$
– Henry Park
May 3 '17 at 12:30
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Compact surfaces form a (commutative) monoid under connected sum where $S^2$ is the identity. If you don't already know that, just consider what connected sum does, it cuts out a disk and identifies the boundary, so if you have anything non-orientable it stays that way, and if you have any handles, it only increases the number of handles.
answered May 3 '17 at 11:59
Adam HughesAdam Hughes
32.2k83770
$endgroup$
$begingroup$
Can you see my edit? Thank you.
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:20
$begingroup$
@RafaelGonzalezLopez as far as your edit goes: yes you can. I didn't appeal to the classification theorem because you didn't indicate whether or not you had already learned it, but with that it is easy as you say.
$endgroup$
– Adam Hughes
May 3 '17 at 13:05
add a comment |
$begingroup$
Compact surfaces form a (commutative) monoid under connected sum where $S^2$ is the identity. If you don't already know that, just consider what connected sum does, it cuts out a disk and identifies the boundary, so if you have anything non-orientable it stays that way, and if you have any handles, it only increases the number of handles.
answered May 3 '17 at 11:59
Adam HughesAdam Hughes
32.2k83770
$endgroup$
$begingroup$
Can you see my edit? Thank you.
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:20
$begingroup$
@RafaelGonzalezLopez as far as your edit goes: yes you can. I didn't appeal to the classification theorem because you didn't indicate whether or not you had already learned it, but with that it is easy as you say.
$endgroup$
– Adam Hughes
May 3 '17 at 13:05
add a comment |
$begingroup$
Compact surfaces form a (commutative) monoid under connected sum where $S^2$ is the identity. If you don't already know that, just consider what connected sum does, it cuts out a disk and identifies the boundary, so if you have anything non-orientable it stays that way, and if you have any handles, it only increases the number of handles.
answered May 3 '17 at 11:59
Adam HughesAdam Hughes
32.2k83770
$endgroup$
Compact surfaces form a (commutative) monoid under connected sum where $S^2$ is the identity. If you don't already know that, just consider what connected sum does, it cuts out a disk and identifies the boundary, so if you have anything non-orientable it stays that way, and if you have any handles, it only increases the number of handles.
answered May 3 '17 at 11:59
Adam HughesAdam Hughes
32.2k83770
answered May 3 '17 at 11:59
Adam HughesAdam Hughes
32.2k83770
answered May 3 '17 at 11:59
Adam HughesAdam Hughes
32.2k83770
answered May 3 '17 at 11:59
Adam HughesAdam Hughes
32.2k83770
32.2k83770
$begingroup$
Can you see my edit? Thank you.
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:20
$begingroup$
@RafaelGonzalezLopez as far as your edit goes: yes you can. I didn't appeal to the classification theorem because you didn't indicate whether or not you had already learned it, but with that it is easy as you say.
$endgroup$
– Adam Hughes
May 3 '17 at 13:05
add a comment |
$begingroup$
Can you see my edit? Thank you.
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:20
$begingroup$
@RafaelGonzalezLopez as far as your edit goes: yes you can. I didn't appeal to the classification theorem because you didn't indicate whether or not you had already learned it, but with that it is easy as you say.
$endgroup$
– Adam Hughes
May 3 '17 at 13:05
$begingroup$
Can you see my edit? Thank you.
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:20
$begingroup$
Can you see my edit? Thank you.
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:20
$begingroup$
@RafaelGonzalezLopez as far as your edit goes: yes you can. I didn't appeal to the classification theorem because you didn't indicate whether or not you had already learned it, but with that it is easy as you say.
$endgroup$
– Adam Hughes
May 3 '17 at 13:05
$begingroup$
@RafaelGonzalezLopez as far as your edit goes: yes you can. I didn't appeal to the classification theorem because you didn't indicate whether or not you had already learned it, but with that it is easy as you say.
$endgroup$
– Adam Hughes
May 3 '17 at 13:05
add a comment |
$begingroup$
As you have mentioned, $A#(B#A#B#cdots)simeq mathbb{R}^2$ and $B#A#B#cdotssimeq mathbb{R}^2$.
But observe that if $X$ is a compact surface, $X#mathbb{R}^2 simeq Xsetminus{x_0}$ (=$X$ minus a point) Therefore, it means that $A$ minus a point is homeomorphic to $mathbb{R}^2$, and then it is easy to conclude that $A simeq S^2$. That $Bsimeq S^2$ follows easily by symmetry.
answered May 3 '17 at 12:06
Henry ParkHenry Park
1,710724
$endgroup$
$begingroup$
That was the point! Thank you for comment, but I would rather prove it without using inifinite sum ^^
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:27
1
$begingroup$
@RafaelGonzalezLopez In fact, it is a very famous argument due to Barry Mazur. Exactly the same argument can be applied to prove some other theorems, such as "the only invertible elements in the monoid of knots and their knot sums is the unknot". See, for instance, math.stonybrook.edu/~jack/MilnorOslo-print.pdf or en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle
$endgroup$
– Henry Park
May 3 '17 at 12:30
add a comment |
$begingroup$
As you have mentioned, $A#(B#A#B#cdots)simeq mathbb{R}^2$ and $B#A#B#cdotssimeq mathbb{R}^2$.
But observe that if $X$ is a compact surface, $X#mathbb{R}^2 simeq Xsetminus{x_0}$ (=$X$ minus a point) Therefore, it means that $A$ minus a point is homeomorphic to $mathbb{R}^2$, and then it is easy to conclude that $A simeq S^2$. That $Bsimeq S^2$ follows easily by symmetry.
answered May 3 '17 at 12:06
Henry ParkHenry Park
1,710724
$endgroup$
$begingroup$
That was the point! Thank you for comment, but I would rather prove it without using inifinite sum ^^
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:27
1
$begingroup$
@RafaelGonzalezLopez In fact, it is a very famous argument due to Barry Mazur. Exactly the same argument can be applied to prove some other theorems, such as "the only invertible elements in the monoid of knots and their knot sums is the unknot". See, for instance, math.stonybrook.edu/~jack/MilnorOslo-print.pdf or en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle
$endgroup$
– Henry Park
May 3 '17 at 12:30
add a comment |
$begingroup$
As you have mentioned, $A#(B#A#B#cdots)simeq mathbb{R}^2$ and $B#A#B#cdotssimeq mathbb{R}^2$.
But observe that if $X$ is a compact surface, $X#mathbb{R}^2 simeq Xsetminus{x_0}$ (=$X$ minus a point) Therefore, it means that $A$ minus a point is homeomorphic to $mathbb{R}^2$, and then it is easy to conclude that $A simeq S^2$. That $Bsimeq S^2$ follows easily by symmetry.
answered May 3 '17 at 12:06
Henry ParkHenry Park
1,710724
$endgroup$
As you have mentioned, $A#(B#A#B#cdots)simeq mathbb{R}^2$ and $B#A#B#cdotssimeq mathbb{R}^2$.
But observe that if $X$ is a compact surface, $X#mathbb{R}^2 simeq Xsetminus{x_0}$ (=$X$ minus a point) Therefore, it means that $A$ minus a point is homeomorphic to $mathbb{R}^2$, and then it is easy to conclude that $A simeq S^2$. That $Bsimeq S^2$ follows easily by symmetry.
answered May 3 '17 at 12:06
Henry ParkHenry Park
1,710724
edited Jan 9 at 21:50
answered May 3 '17 at 12:06
Henry ParkHenry Park
1,710724
answered May 3 '17 at 12:06
Henry ParkHenry Park
1,710724
answered May 3 '17 at 12:06
Henry ParkHenry Park
1,710724
1,710724
$begingroup$
That was the point! Thank you for comment, but I would rather prove it without using inifinite sum ^^
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:27
1
$begingroup$
@RafaelGonzalezLopez In fact, it is a very famous argument due to Barry Mazur. Exactly the same argument can be applied to prove some other theorems, such as "the only invertible elements in the monoid of knots and their knot sums is the unknot". See, for instance, math.stonybrook.edu/~jack/MilnorOslo-print.pdf or en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle
$endgroup$
– Henry Park
May 3 '17 at 12:30
add a comment |
$begingroup$
That was the point! Thank you for comment, but I would rather prove it without using inifinite sum ^^
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:27
1
$begingroup$
@RafaelGonzalezLopez In fact, it is a very famous argument due to Barry Mazur. Exactly the same argument can be applied to prove some other theorems, such as "the only invertible elements in the monoid of knots and their knot sums is the unknot". See, for instance, math.stonybrook.edu/~jack/MilnorOslo-print.pdf or en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle
$endgroup$
– Henry Park
May 3 '17 at 12:30
$begingroup$
That was the point! Thank you for comment, but I would rather prove it without using inifinite sum ^^
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:27
$begingroup$
That was the point! Thank you for comment, but I would rather prove it without using inifinite sum ^^
$endgroup$
– Rafael Gonzalez Lopez
May 3 '17 at 12:27
1
1
$begingroup$
@RafaelGonzalezLopez In fact, it is a very famous argument due to Barry Mazur. Exactly the same argument can be applied to prove some other theorems, such as "the only invertible elements in the monoid of knots and their knot sums is the unknot". See, for instance, math.stonybrook.edu/~jack/MilnorOslo-print.pdf or en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle
$endgroup$
– Henry Park
May 3 '17 at 12:30
$begingroup$
@RafaelGonzalezLopez In fact, it is a very famous argument due to Barry Mazur. Exactly the same argument can be applied to prove some other theorems, such as "the only invertible elements in the monoid of knots and their knot sums is the unknot". See, for instance, math.stonybrook.edu/~jack/MilnorOslo-print.pdf or en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle
$endgroup$
– Henry Park
May 3 '17 at 12:30
add a comment |
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