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
1






active

oldest

votes








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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How do I know what Microsoft account the skydrive app is syncing to?

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up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
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When does type information flow backwards in C++?

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up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
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Grease: Live!

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A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
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