How to show that $Bbb C/(-infty ,0]$ is simply connected space












0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You could show it's homeomorphic to $Bbb C$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 26 '18 at 18:29










  • $begingroup$
    we cannot show it by homeomorphism, sorry...
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:30






  • 1




    $begingroup$
    It's not convex, but still star-shaped.
    $endgroup$
    – Jacky Chong
    Dec 26 '18 at 18:31








  • 4




    $begingroup$
    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.
    $endgroup$
    – greelious
    Dec 26 '18 at 18:32










  • $begingroup$
    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?
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:59


















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    You could show it's homeomorphic to $Bbb C$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 26 '18 at 18:29










  • $begingroup$
    we cannot show it by homeomorphism, sorry...
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:30






  • 1




    $begingroup$
    It's not convex, but still star-shaped.
    $endgroup$
    – Jacky Chong
    Dec 26 '18 at 18:31








  • 4




    $begingroup$
    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.
    $endgroup$
    – greelious
    Dec 26 '18 at 18:32










  • $begingroup$
    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?
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:59
















0












0








0


0



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 26 '18 at 18:27









Daniel VainshteinDaniel Vainshtein

19011




19011








  • 1




    $begingroup$
    You could show it's homeomorphic to $Bbb C$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 26 '18 at 18:29










  • $begingroup$
    we cannot show it by homeomorphism, sorry...
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:30






  • 1




    $begingroup$
    It's not convex, but still star-shaped.
    $endgroup$
    – Jacky Chong
    Dec 26 '18 at 18:31








  • 4




    $begingroup$
    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.
    $endgroup$
    – greelious
    Dec 26 '18 at 18:32










  • $begingroup$
    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?
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:59
















  • 1




    $begingroup$
    You could show it's homeomorphic to $Bbb C$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 26 '18 at 18:29










  • $begingroup$
    we cannot show it by homeomorphism, sorry...
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:30






  • 1




    $begingroup$
    It's not convex, but still star-shaped.
    $endgroup$
    – Jacky Chong
    Dec 26 '18 at 18:31








  • 4




    $begingroup$
    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.
    $endgroup$
    – greelious
    Dec 26 '18 at 18:32










  • $begingroup$
    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?
    $endgroup$
    – Daniel Vainshtein
    Dec 26 '18 at 18:59










1




1




$begingroup$
You could show it's homeomorphic to $Bbb C$.
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 18:29




$begingroup$
You could show it's homeomorphic to $Bbb C$.
$endgroup$
– Lord Shark the Unknown
Dec 26 '18 at 18:29












$begingroup$
we cannot show it by homeomorphism, sorry...
$endgroup$
– Daniel Vainshtein
Dec 26 '18 at 18:30




$begingroup$
we cannot show it by homeomorphism, sorry...
$endgroup$
– Daniel Vainshtein
Dec 26 '18 at 18:30




1




1




$begingroup$
It's not convex, but still star-shaped.
$endgroup$
– Jacky Chong
Dec 26 '18 at 18:31






$begingroup$
It's not convex, but still star-shaped.
$endgroup$
– Jacky Chong
Dec 26 '18 at 18:31






4




4




$begingroup$
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.
$endgroup$
– greelious
Dec 26 '18 at 18:32




$begingroup$
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.
$endgroup$
– greelious
Dec 26 '18 at 18:32












$begingroup$
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?
$endgroup$
– Daniel Vainshtein
Dec 26 '18 at 18:59






$begingroup$
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?
$endgroup$
– Daniel Vainshtein
Dec 26 '18 at 18:59












1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    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 $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
    $endgroup$
    – ecrin
    Dec 26 '18 at 19:09












  • $begingroup$
    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?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 8:57












  • $begingroup$
    No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
    $endgroup$
    – ecrin
    Dec 27 '18 at 9:48










  • $begingroup$
    so how do I find a literal homotopy between 2 pathes with the same start and end?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 20:32










  • $begingroup$
    I showed it in the previous comment, you have to use polar coordinates
    $endgroup$
    – ecrin
    Dec 27 '18 at 22:14











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    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 $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
    $endgroup$
    – ecrin
    Dec 26 '18 at 19:09












  • $begingroup$
    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?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 8:57












  • $begingroup$
    No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
    $endgroup$
    – ecrin
    Dec 27 '18 at 9:48










  • $begingroup$
    so how do I find a literal homotopy between 2 pathes with the same start and end?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 20:32










  • $begingroup$
    I showed it in the previous comment, you have to use polar coordinates
    $endgroup$
    – ecrin
    Dec 27 '18 at 22:14
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    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 $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
    $endgroup$
    – ecrin
    Dec 26 '18 at 19:09












  • $begingroup$
    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?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 8:57












  • $begingroup$
    No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
    $endgroup$
    – ecrin
    Dec 27 '18 at 9:48










  • $begingroup$
    so how do I find a literal homotopy between 2 pathes with the same start and end?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 20:32










  • $begingroup$
    I showed it in the previous comment, you have to use polar coordinates
    $endgroup$
    – ecrin
    Dec 27 '18 at 22:14














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 '18 at 18:50









ecrinecrin

3477




3477








  • 1




    $begingroup$
    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 $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
    $endgroup$
    – ecrin
    Dec 26 '18 at 19:09












  • $begingroup$
    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?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 8:57












  • $begingroup$
    No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
    $endgroup$
    – ecrin
    Dec 27 '18 at 9:48










  • $begingroup$
    so how do I find a literal homotopy between 2 pathes with the same start and end?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 20:32










  • $begingroup$
    I showed it in the previous comment, you have to use polar coordinates
    $endgroup$
    – ecrin
    Dec 27 '18 at 22:14














  • 1




    $begingroup$
    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 $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
    $endgroup$
    – ecrin
    Dec 26 '18 at 19:09












  • $begingroup$
    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?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 8:57












  • $begingroup$
    No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
    $endgroup$
    – ecrin
    Dec 27 '18 at 9:48










  • $begingroup$
    so how do I find a literal homotopy between 2 pathes with the same start and end?
    $endgroup$
    – Daniel Vainshtein
    Dec 27 '18 at 20:32










  • $begingroup$
    I showed it in the previous comment, you have to use polar coordinates
    $endgroup$
    – ecrin
    Dec 27 '18 at 22:14








1




1




$begingroup$
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 $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
$endgroup$
– ecrin
Dec 26 '18 at 19:09






$begingroup$
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 $H(s,t)=(1-s)(gamma_{rho}(t),gamma_{theta}(t))+s(eta_{rho}(t),eta_{theta}(t))$.
$endgroup$
– ecrin
Dec 26 '18 at 19:09














$begingroup$
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?
$endgroup$
– Daniel Vainshtein
Dec 27 '18 at 8:57






$begingroup$
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?
$endgroup$
– Daniel Vainshtein
Dec 27 '18 at 8:57














$begingroup$
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
$endgroup$
– ecrin
Dec 27 '18 at 9:48




$begingroup$
No, it's not. In fact your homotopy isn't rel{0,1}: $H(s,0)=gamma(0)*(1-2s)+2snegamma(0)$ for $sne 0$.
$endgroup$
– ecrin
Dec 27 '18 at 9:48












$begingroup$
so how do I find a literal homotopy between 2 pathes with the same start and end?
$endgroup$
– Daniel Vainshtein
Dec 27 '18 at 20:32




$begingroup$
so how do I find a literal homotopy between 2 pathes with the same start and end?
$endgroup$
– Daniel Vainshtein
Dec 27 '18 at 20:32












$begingroup$
I showed it in the previous comment, you have to use polar coordinates
$endgroup$
– ecrin
Dec 27 '18 at 22:14




$begingroup$
I showed it in the previous comment, you have to use polar coordinates
$endgroup$
– ecrin
Dec 27 '18 at 22:14


















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Probability when a professor distributes a quiz and homework assignment to a class of n students.

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0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
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Aardman Animations

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Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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Are they similar matrix

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0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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