$mathbb C P^1cong (D^2times{1}+ D^2times {-1})big /_sim $
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I need to proove a bigger result that $mathbb C P^1$ is homeomorphic to $S^2$.
For that I have already showed $$S^2cong (D^2times{1}+ D^2times {-1})big /_sim$$ where $(z,1)sim (z,-1) iff z in S^1$ and identity anywhere else.
Now I'm searching for two subsets in $mathbb C P^1$ which union is homeomorphic to $D^2times{1}+D^2times {-1}big /_sim$.
I tried to partition lines $[a:b]$ in $mathbb C^2$ in such sets where either $[a:b]=[1:c]$ or $[a:b]=[0:d]$, but second one is to "small" to be homeomorphic.
I need some hints guys.
general-topology algebraic-topology homotopy-theory projective-space
asked Nov 12 at 20:37
user3342072
373113
add a comment |
up vote
1
down vote
favorite
I need to proove a bigger result that $mathbb C P^1$ is homeomorphic to $S^2$.
For that I have already showed $$S^2cong (D^2times{1}+ D^2times {-1})big /_sim$$ where $(z,1)sim (z,-1) iff z in S^1$ and identity anywhere else.
Now I'm searching for two subsets in $mathbb C P^1$ which union is homeomorphic to $D^2times{1}+D^2times {-1}big /_sim$.
I tried to partition lines $[a:b]$ in $mathbb C^2$ in such sets where either $[a:b]=[1:c]$ or $[a:b]=[0:d]$, but second one is to "small" to be homeomorphic.
I need some hints guys.
general-topology algebraic-topology homotopy-theory projective-space
asked Nov 12 at 20:37
user3342072
373113
1
First, we may assume $|a|^2 + |b|^2 = 1$ (or, depending on your definition of $mathbb{C}P^n$, this may automatically be true). Now...Hint: Think about $|a|^2 leq 1/2$ and $|a|^2geq 1/2$ separately.
– Jason DeVito
Nov 12 at 20:43
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to proove a bigger result that $mathbb C P^1$ is homeomorphic to $S^2$.
For that I have already showed $$S^2cong (D^2times{1}+ D^2times {-1})big /_sim$$ where $(z,1)sim (z,-1) iff z in S^1$ and identity anywhere else.
Now I'm searching for two subsets in $mathbb C P^1$ which union is homeomorphic to $D^2times{1}+D^2times {-1}big /_sim$.
I tried to partition lines $[a:b]$ in $mathbb C^2$ in such sets where either $[a:b]=[1:c]$ or $[a:b]=[0:d]$, but second one is to "small" to be homeomorphic.
I need some hints guys.
general-topology algebraic-topology homotopy-theory projective-space
asked Nov 12 at 20:37
user3342072
373113
I need to proove a bigger result that $mathbb C P^1$ is homeomorphic to $S^2$.
For that I have already showed $$S^2cong (D^2times{1}+ D^2times {-1})big /_sim$$ where $(z,1)sim (z,-1) iff z in S^1$ and identity anywhere else.
Now I'm searching for two subsets in $mathbb C P^1$ which union is homeomorphic to $D^2times{1}+D^2times {-1}big /_sim$.
I tried to partition lines $[a:b]$ in $mathbb C^2$ in such sets where either $[a:b]=[1:c]$ or $[a:b]=[0:d]$, but second one is to "small" to be homeomorphic.
I need some hints guys.
general-topology algebraic-topology homotopy-theory projective-space
general-topology algebraic-topology homotopy-theory projective-space
asked Nov 12 at 20:37
user3342072
373113
asked Nov 12 at 20:37
user3342072
373113
asked Nov 12 at 20:37
user3342072
373113
asked Nov 12 at 20:37
user3342072
373113
asked Nov 12 at 20:37
user3342072
373113
373113
1
First, we may assume $|a|^2 + |b|^2 = 1$ (or, depending on your definition of $mathbb{C}P^n$, this may automatically be true). Now...Hint: Think about $|a|^2 leq 1/2$ and $|a|^2geq 1/2$ separately.
– Jason DeVito
Nov 12 at 20:43
add a comment |
1
First, we may assume $|a|^2 + |b|^2 = 1$ (or, depending on your definition of $mathbb{C}P^n$, this may automatically be true). Now...Hint: Think about $|a|^2 leq 1/2$ and $|a|^2geq 1/2$ separately.
– Jason DeVito
Nov 12 at 20:43
1
1
First, we may assume $|a|^2 + |b|^2 = 1$ (or, depending on your definition of $mathbb{C}P^n$, this may automatically be true). Now...Hint: Think about $|a|^2 leq 1/2$ and $|a|^2geq 1/2$ separately.
– Jason DeVito
Nov 12 at 20:43
First, we may assume $|a|^2 + |b|^2 = 1$ (or, depending on your definition of $mathbb{C}P^n$, this may automatically be true). Now...Hint: Think about $|a|^2 leq 1/2$ and $|a|^2geq 1/2$ separately.
– Jason DeVito
Nov 12 at 20:43
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Your method does work.
Take $$ D^2_{rm first} = { [1 : z] in mathbb {CP}^1 : | z | leq 1 }$$ $$ D^2_{rm second} = { [w : 1] in mathbb {CP}^1 : | w | leq 1 }$$
and identify the boundary points like this:
$$[1 :e^{itheta}] in D^2_{rm first} sim [ e^{-itheta} : 1] in D^2_{rm second} {rm for all } theta in [0, 2pi) $$
It is then immediate that
$$ mathbb {CP}^1 cong left( D^2_{rm first} coprod D^2_{rm second} right) / sim cong S^2. $$
answered Nov 12 at 20:57
Kenny Wong
16.6k21135
I admire your answer. Although it appears that you have directly identified $D^2_{rm first}subset mathbb C P^1$ with $D^2subset mathbb R^2$, but that's okay I guess? I think I got it, thank you!
– user3342072
Nov 12 at 22:41
1
Yes, we can say that $mathbb {CP}^1 = D_{rm first}^2 cup D_{rm second}^2$. But by introducing the equivalence relation, we are spelling out how the two disks are joined together, and this is important. And yes, I'm identifying $D_{rm first}^2 subset mathbb {CP}^1$ with the disk $D^2 := { z in mathbb C : |z | leq 1 } subset mathbb C$, which is obvious homeomorphic to the disk $D^2 := { (x, y) in mathbb R^2 : sqrt{x^2 + y^2} leq 1 } subset mathbb R^2$.
– Kenny Wong
Nov 12 at 22:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your method does work.
Take $$ D^2_{rm first} = { [1 : z] in mathbb {CP}^1 : | z | leq 1 }$$ $$ D^2_{rm second} = { [w : 1] in mathbb {CP}^1 : | w | leq 1 }$$
and identify the boundary points like this:
$$[1 :e^{itheta}] in D^2_{rm first} sim [ e^{-itheta} : 1] in D^2_{rm second} {rm for all } theta in [0, 2pi) $$
It is then immediate that
$$ mathbb {CP}^1 cong left( D^2_{rm first} coprod D^2_{rm second} right) / sim cong S^2. $$
answered Nov 12 at 20:57
Kenny Wong
16.6k21135
I admire your answer. Although it appears that you have directly identified $D^2_{rm first}subset mathbb C P^1$ with $D^2subset mathbb R^2$, but that's okay I guess? I think I got it, thank you!
– user3342072
Nov 12 at 22:41
1
Yes, we can say that $mathbb {CP}^1 = D_{rm first}^2 cup D_{rm second}^2$. But by introducing the equivalence relation, we are spelling out how the two disks are joined together, and this is important. And yes, I'm identifying $D_{rm first}^2 subset mathbb {CP}^1$ with the disk $D^2 := { z in mathbb C : |z | leq 1 } subset mathbb C$, which is obvious homeomorphic to the disk $D^2 := { (x, y) in mathbb R^2 : sqrt{x^2 + y^2} leq 1 } subset mathbb R^2$.
– Kenny Wong
Nov 12 at 22:44
add a comment |
up vote
1
down vote
accepted
Your method does work.
Take $$ D^2_{rm first} = { [1 : z] in mathbb {CP}^1 : | z | leq 1 }$$ $$ D^2_{rm second} = { [w : 1] in mathbb {CP}^1 : | w | leq 1 }$$
and identify the boundary points like this:
$$[1 :e^{itheta}] in D^2_{rm first} sim [ e^{-itheta} : 1] in D^2_{rm second} {rm for all } theta in [0, 2pi) $$
It is then immediate that
$$ mathbb {CP}^1 cong left( D^2_{rm first} coprod D^2_{rm second} right) / sim cong S^2. $$
answered Nov 12 at 20:57
Kenny Wong
16.6k21135
I admire your answer. Although it appears that you have directly identified $D^2_{rm first}subset mathbb C P^1$ with $D^2subset mathbb R^2$, but that's okay I guess? I think I got it, thank you!
– user3342072
Nov 12 at 22:41
1
Yes, we can say that $mathbb {CP}^1 = D_{rm first}^2 cup D_{rm second}^2$. But by introducing the equivalence relation, we are spelling out how the two disks are joined together, and this is important. And yes, I'm identifying $D_{rm first}^2 subset mathbb {CP}^1$ with the disk $D^2 := { z in mathbb C : |z | leq 1 } subset mathbb C$, which is obvious homeomorphic to the disk $D^2 := { (x, y) in mathbb R^2 : sqrt{x^2 + y^2} leq 1 } subset mathbb R^2$.
– Kenny Wong
Nov 12 at 22:44
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your method does work.
Take $$ D^2_{rm first} = { [1 : z] in mathbb {CP}^1 : | z | leq 1 }$$ $$ D^2_{rm second} = { [w : 1] in mathbb {CP}^1 : | w | leq 1 }$$
and identify the boundary points like this:
$$[1 :e^{itheta}] in D^2_{rm first} sim [ e^{-itheta} : 1] in D^2_{rm second} {rm for all } theta in [0, 2pi) $$
It is then immediate that
$$ mathbb {CP}^1 cong left( D^2_{rm first} coprod D^2_{rm second} right) / sim cong S^2. $$
answered Nov 12 at 20:57
Kenny Wong
16.6k21135
Your method does work.
Take $$ D^2_{rm first} = { [1 : z] in mathbb {CP}^1 : | z | leq 1 }$$ $$ D^2_{rm second} = { [w : 1] in mathbb {CP}^1 : | w | leq 1 }$$
and identify the boundary points like this:
$$[1 :e^{itheta}] in D^2_{rm first} sim [ e^{-itheta} : 1] in D^2_{rm second} {rm for all } theta in [0, 2pi) $$
It is then immediate that
$$ mathbb {CP}^1 cong left( D^2_{rm first} coprod D^2_{rm second} right) / sim cong S^2. $$
answered Nov 12 at 20:57
Kenny Wong
16.6k21135
answered Nov 12 at 20:57
Kenny Wong
16.6k21135
answered Nov 12 at 20:57
Kenny Wong
16.6k21135
answered Nov 12 at 20:57
Kenny Wong
16.6k21135
16.6k21135
I admire your answer. Although it appears that you have directly identified $D^2_{rm first}subset mathbb C P^1$ with $D^2subset mathbb R^2$, but that's okay I guess? I think I got it, thank you!
– user3342072
Nov 12 at 22:41
1
Yes, we can say that $mathbb {CP}^1 = D_{rm first}^2 cup D_{rm second}^2$. But by introducing the equivalence relation, we are spelling out how the two disks are joined together, and this is important. And yes, I'm identifying $D_{rm first}^2 subset mathbb {CP}^1$ with the disk $D^2 := { z in mathbb C : |z | leq 1 } subset mathbb C$, which is obvious homeomorphic to the disk $D^2 := { (x, y) in mathbb R^2 : sqrt{x^2 + y^2} leq 1 } subset mathbb R^2$.
– Kenny Wong
Nov 12 at 22:44
add a comment |
I admire your answer. Although it appears that you have directly identified $D^2_{rm first}subset mathbb C P^1$ with $D^2subset mathbb R^2$, but that's okay I guess? I think I got it, thank you!
– user3342072
Nov 12 at 22:41
1
Yes, we can say that $mathbb {CP}^1 = D_{rm first}^2 cup D_{rm second}^2$. But by introducing the equivalence relation, we are spelling out how the two disks are joined together, and this is important. And yes, I'm identifying $D_{rm first}^2 subset mathbb {CP}^1$ with the disk $D^2 := { z in mathbb C : |z | leq 1 } subset mathbb C$, which is obvious homeomorphic to the disk $D^2 := { (x, y) in mathbb R^2 : sqrt{x^2 + y^2} leq 1 } subset mathbb R^2$.
– Kenny Wong
Nov 12 at 22:44
I admire your answer. Although it appears that you have directly identified $D^2_{rm first}subset mathbb C P^1$ with $D^2subset mathbb R^2$, but that's okay I guess? I think I got it, thank you!
– user3342072
Nov 12 at 22:41
I admire your answer. Although it appears that you have directly identified $D^2_{rm first}subset mathbb C P^1$ with $D^2subset mathbb R^2$, but that's okay I guess? I think I got it, thank you!
– user3342072
Nov 12 at 22:41
1
1
Yes, we can say that $mathbb {CP}^1 = D_{rm first}^2 cup D_{rm second}^2$. But by introducing the equivalence relation, we are spelling out how the two disks are joined together, and this is important. And yes, I'm identifying $D_{rm first}^2 subset mathbb {CP}^1$ with the disk $D^2 := { z in mathbb C : |z | leq 1 } subset mathbb C$, which is obvious homeomorphic to the disk $D^2 := { (x, y) in mathbb R^2 : sqrt{x^2 + y^2} leq 1 } subset mathbb R^2$.
– Kenny Wong
Nov 12 at 22:44
Yes, we can say that $mathbb {CP}^1 = D_{rm first}^2 cup D_{rm second}^2$. But by introducing the equivalence relation, we are spelling out how the two disks are joined together, and this is important. And yes, I'm identifying $D_{rm first}^2 subset mathbb {CP}^1$ with the disk $D^2 := { z in mathbb C : |z | leq 1 } subset mathbb C$, which is obvious homeomorphic to the disk $D^2 := { (x, y) in mathbb R^2 : sqrt{x^2 + y^2} leq 1 } subset mathbb R^2$.
– Kenny Wong
Nov 12 at 22:44
add a comment |
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First, we may assume $|a|^2 + |b|^2 = 1$ (or, depending on your definition of $mathbb{C}P^n$, this may automatically be true). Now...Hint: Think about $|a|^2 leq 1/2$ and $|a|^2geq 1/2$ separately.
– Jason DeVito
Nov 12 at 20:43