How to show that $Bbb C/(-infty ,0]$ is simply connected space
I need to show that for $Bbb C/(-infty ,0]$ that if I have arbitrary different pathes $gamma , eta :[0,1] toBbb C/(-infty ,0]$ such that $gamma(0)=eta(0)=alpha $ and $gamma(1)=eta(1)=beta $ I have continuous homotopy $H:[0,1]^2 to Bbb C/(-infty ,0]$ such that: $H(0,t) = gamma (t),H(1,t) = eta (t),H(s,0) = alpha,H(s,1) = beta$
unfortunatily my space isn`t a convex,so how do I define a proper Homotopy and proving it's continuous?
general-topology complex-analysis
asked Dec 26 '18 at 18:27
Daniel Vainshtein
18811
add a comment |
I need to show that for $Bbb C/(-infty ,0]$ that if I have arbitrary different pathes $gamma , eta :[0,1] toBbb C/(-infty ,0]$ such that $gamma(0)=eta(0)=alpha $ and $gamma(1)=eta(1)=beta $ I have continuous homotopy $H:[0,1]^2 to Bbb C/(-infty ,0]$ such that: $H(0,t) = gamma (t),H(1,t) = eta (t),H(s,0) = alpha,H(s,1) = beta$
unfortunatily my space isn`t a convex,so how do I define a proper Homotopy and proving it's continuous?
general-topology complex-analysis
asked Dec 26 '18 at 18:27
Daniel Vainshtein
18811
1
You could show it's homeomorphic to $Bbb C$.
– Lord Shark the Unknown
Dec 26 '18 at 18:29
we cannot show it by homeomorphism, sorry...
– Daniel Vainshtein
Dec 26 '18 at 18:30
1
It's not convex, but still star-shaped.
– Jacky Chong
Dec 26 '18 at 18:31
4
Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
– greelious
Dec 26 '18 at 18:32
Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
– Daniel Vainshtein
Dec 26 '18 at 18:59
add a comment |
I need to show that for $Bbb C/(-infty ,0]$ that if I have arbitrary different pathes $gamma , eta :[0,1] toBbb C/(-infty ,0]$ such that $gamma(0)=eta(0)=alpha $ and $gamma(1)=eta(1)=beta $ I have continuous homotopy $H:[0,1]^2 to Bbb C/(-infty ,0]$ such that: $H(0,t) = gamma (t),H(1,t) = eta (t),H(s,0) = alpha,H(s,1) = beta$
unfortunatily my space isn`t a convex,so how do I define a proper Homotopy and proving it's continuous?
general-topology complex-analysis
asked Dec 26 '18 at 18:27
Daniel Vainshtein
18811
I need to show that for $Bbb C/(-infty ,0]$ that if I have arbitrary different pathes $gamma , eta :[0,1] toBbb C/(-infty ,0]$ such that $gamma(0)=eta(0)=alpha $ and $gamma(1)=eta(1)=beta $ I have continuous homotopy $H:[0,1]^2 to Bbb C/(-infty ,0]$ such that: $H(0,t) = gamma (t),H(1,t) = eta (t),H(s,0) = alpha,H(s,1) = beta$
unfortunatily my space isn`t a convex,so how do I define a proper Homotopy and proving it's continuous?
general-topology complex-analysis
general-topology complex-analysis
asked Dec 26 '18 at 18:27
Daniel Vainshtein
18811
asked Dec 26 '18 at 18:27
Daniel Vainshtein
18811
asked Dec 26 '18 at 18:27
Daniel Vainshtein
18811
asked Dec 26 '18 at 18:27
Daniel Vainshtein
18811
asked Dec 26 '18 at 18:27
Daniel Vainshtein
18811
18811
1
You could show it's homeomorphic to $Bbb C$.
– Lord Shark the Unknown
Dec 26 '18 at 18:29
we cannot show it by homeomorphism, sorry...
– Daniel Vainshtein
Dec 26 '18 at 18:30
1
It's not convex, but still star-shaped.
– Jacky Chong
Dec 26 '18 at 18:31
4
Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
– greelious
Dec 26 '18 at 18:32
Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
– Daniel Vainshtein
Dec 26 '18 at 18:59
add a comment |
1
You could show it's homeomorphic to $Bbb C$.
– Lord Shark the Unknown
Dec 26 '18 at 18:29
we cannot show it by homeomorphism, sorry...
– Daniel Vainshtein
Dec 26 '18 at 18:30
1
It's not convex, but still star-shaped.
– Jacky Chong
Dec 26 '18 at 18:31
4
Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
– greelious
Dec 26 '18 at 18:32
Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
– Daniel Vainshtein
Dec 26 '18 at 18:59
1
1
You could show it's homeomorphic to $Bbb C$.
– Lord Shark the Unknown
Dec 26 '18 at 18:29
You could show it's homeomorphic to $Bbb C$.
– Lord Shark the Unknown
Dec 26 '18 at 18:29
we cannot show it by homeomorphism, sorry...
– Daniel Vainshtein
Dec 26 '18 at 18:30
we cannot show it by homeomorphism, sorry...
– Daniel Vainshtein
Dec 26 '18 at 18:30
1
1
It's not convex, but still star-shaped.
– Jacky Chong
Dec 26 '18 at 18:31
It's not convex, but still star-shaped.
– Jacky Chong
Dec 26 '18 at 18:31
4
4
Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
– greelious
Dec 26 '18 at 18:32
Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
– greelious
Dec 26 '18 at 18:32
Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
– Daniel Vainshtein
Dec 26 '18 at 18:59
Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
– Daniel Vainshtein
Dec 26 '18 at 18:59
add a comment |
1 Answer
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votes
Consider the complex plane using polar coordinate system.
Every points in $mathbb{C}setminus(-inf,0]$ can be written as $(rho,theta)$, $rhoin(0,+inf)$, $thetain(-pi,pi)$.
The function $fcolonmathbb{C}setminus(-inf,0]to{zinmathbb{C}, Im(z)>0}$ defined as $f(rho,theta)=(rho,frac{theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.
answered Dec 26 '18 at 18:50
ecrin
714
New contributor
ecrin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
– ecrin
Dec 26 '18 at 19:09
Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
– Daniel Vainshtein
Dec 27 '18 at 8:57
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
– ecrin
Dec 27 '18 at 9:48
so how do I find a literal homotopy between 2 pathes with the same start and end?
– Daniel Vainshtein
Dec 27 '18 at 20:32
I showed it in the previous comment, you have to use polar coordinates
– ecrin
Dec 27 '18 at 22:14
add a comment |
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Consider the complex plane using polar coordinate system.
Every points in $mathbb{C}setminus(-inf,0]$ can be written as $(rho,theta)$, $rhoin(0,+inf)$, $thetain(-pi,pi)$.
The function $fcolonmathbb{C}setminus(-inf,0]to{zinmathbb{C}, Im(z)>0}$ defined as $f(rho,theta)=(rho,frac{theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.
answered Dec 26 '18 at 18:50
ecrin
714
New contributor
ecrin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
– ecrin
Dec 26 '18 at 19:09
Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
– Daniel Vainshtein
Dec 27 '18 at 8:57
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
– ecrin
Dec 27 '18 at 9:48
so how do I find a literal homotopy between 2 pathes with the same start and end?
– Daniel Vainshtein
Dec 27 '18 at 20:32
I showed it in the previous comment, you have to use polar coordinates
– ecrin
Dec 27 '18 at 22:14
add a comment |
Consider the complex plane using polar coordinate system.
Every points in $mathbb{C}setminus(-inf,0]$ can be written as $(rho,theta)$, $rhoin(0,+inf)$, $thetain(-pi,pi)$.
The function $fcolonmathbb{C}setminus(-inf,0]to{zinmathbb{C}, Im(z)>0}$ defined as $f(rho,theta)=(rho,frac{theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.
answered Dec 26 '18 at 18:50
ecrin
714
New contributor
ecrin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
– ecrin
Dec 26 '18 at 19:09
Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
– Daniel Vainshtein
Dec 27 '18 at 8:57
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
– ecrin
Dec 27 '18 at 9:48
so how do I find a literal homotopy between 2 pathes with the same start and end?
– Daniel Vainshtein
Dec 27 '18 at 20:32
I showed it in the previous comment, you have to use polar coordinates
– ecrin
Dec 27 '18 at 22:14
add a comment |
Consider the complex plane using polar coordinate system.
Every points in $mathbb{C}setminus(-inf,0]$ can be written as $(rho,theta)$, $rhoin(0,+inf)$, $thetain(-pi,pi)$.
The function $fcolonmathbb{C}setminus(-inf,0]to{zinmathbb{C}, Im(z)>0}$ defined as $f(rho,theta)=(rho,frac{theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.
answered Dec 26 '18 at 18:50
ecrin
714
New contributor
ecrin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider the complex plane using polar coordinate system.
Every points in $mathbb{C}setminus(-inf,0]$ can be written as $(rho,theta)$, $rhoin(0,+inf)$, $thetain(-pi,pi)$.
The function $fcolonmathbb{C}setminus(-inf,0]to{zinmathbb{C}, Im(z)>0}$ defined as $f(rho,theta)=(rho,frac{theta}{2})$ is an homeomorphism, and Imf is a convess space, so the domain is simply connected.
answered Dec 26 '18 at 18:50
ecrin
714
New contributor
ecrin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 26 '18 at 18:50
ecrin
714
New contributor
ecrin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 26 '18 at 18:50
ecrin
714
answered Dec 26 '18 at 18:50
ecrin
714
714
New contributor
ecrin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ecrin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ecrin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
– ecrin
Dec 26 '18 at 19:09
Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
– Daniel Vainshtein
Dec 27 '18 at 8:57
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
– ecrin
Dec 27 '18 at 9:48
so how do I find a literal homotopy between 2 pathes with the same start and end?
– Daniel Vainshtein
Dec 27 '18 at 20:32
I showed it in the previous comment, you have to use polar coordinates
– ecrin
Dec 27 '18 at 22:14
add a comment |
1
I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
– ecrin
Dec 26 '18 at 19:09
Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
– Daniel Vainshtein
Dec 27 '18 at 8:57
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
– ecrin
Dec 27 '18 at 9:48
so how do I find a literal homotopy between 2 pathes with the same start and end?
– Daniel Vainshtein
Dec 27 '18 at 20:32
I showed it in the previous comment, you have to use polar coordinates
– ecrin
Dec 27 '18 at 22:14
1
1
I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
– ecrin
Dec 26 '18 at 19:09
I noticed that you're looking for an explicit homotopy. The homeomorphism helps you in this search: write the two curves in polar coordinates as $gamma(t)=(gamma_{rho}(t),gamma_{theta}(t))$ and $eta(t)=(eta_{rho}(t),eta_{theta}(t))$, then the homotopy is $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
– ecrin
Dec 26 '18 at 19:09
Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
– Daniel Vainshtein
Dec 27 '18 at 8:57
Iv'e tried to build homotopy such that: $H(s,t)=γ(t)∗(1−2s)+2s,0≤s≤0.5$ and $(2−2s)+η(t)(2s−1),0.5≤s≤1$ does it`s ok?
– Daniel Vainshtein
Dec 27 '18 at 8:57
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
– ecrin
Dec 27 '18 at 9:48
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
– ecrin
Dec 27 '18 at 9:48
so how do I find a literal homotopy between 2 pathes with the same start and end?
– Daniel Vainshtein
Dec 27 '18 at 20:32
so how do I find a literal homotopy between 2 pathes with the same start and end?
– Daniel Vainshtein
Dec 27 '18 at 20:32
I showed it in the previous comment, you have to use polar coordinates
– ecrin
Dec 27 '18 at 22:14
I showed it in the previous comment, you have to use polar coordinates
– ecrin
Dec 27 '18 at 22:14
add a comment |
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You could show it's homeomorphic to $Bbb C$.
– Lord Shark the Unknown
Dec 26 '18 at 18:29
we cannot show it by homeomorphism, sorry...
– Daniel Vainshtein
Dec 26 '18 at 18:30
1
It's not convex, but still star-shaped.
– Jacky Chong
Dec 26 '18 at 18:31
4
Note that $Bbb C/(-infty ,0]$ is star-shaped with center $1$ for example. You could use the straight line homotopy on each of the paths to obtain homotopies to the center, then concatenate one with the reverse of the other and use the Pasting lemma to show continuity.
– greelious
Dec 26 '18 at 18:32
Iv'e tried to build homotopy such that: $H(s,t)=gamma(t)*(1-2s)+2s , 0le s le 0.5$ and $(2-2s)+eta(t)(2s-1) ,0.5 le s le 1$ does it`s ok?
– Daniel Vainshtein
Dec 26 '18 at 18:59