Czes Kosniowski: Homeomorphic topological spaces $S^1times I / sim$ and $D^2$ [duplicate]$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$Proof that $mathbb R/[0,1]$ and $mathbb R$ are homeomorphicShow $(S^1times [0,1])/$~ is homeomorphic to $D^2$Topological , Homeomorphic version of $|S times S|=|S| $Why is $D^n/sim$ homeomorphic to $mathbbRP^n$?Why $(mathbb Qtimesmathbb Q)/(mathbb Ztimes=)$ is not homeomorphic to $(mathbb Q/mathbb Z)times(mathbb Q/=)$?Quotient spaces homeomorphic to the realsProve that $(Xtimes [0,1])/sim'$ is homeomorphic to $(X/sim)times[0,1].$RP2 is homeomorphic to $(I times I)/ sim$Homeomorphism between a quotient topology on $mathbbR^2$ and $mathbbR$

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Czes Kosniowski: Homeomorphic topological spaces $S^1times I / sim$ and $D^2$ [duplicate]


$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$$(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$Proof that $mathbb R/[0,1]$ and $mathbb R$ are homeomorphicShow $(S^1times [0,1])/$~ is homeomorphic to $D^2$Topological , Homeomorphic version of $|S times S|=|S| $Why is $D^n/sim$ homeomorphic to $mathbbRP^n$?Why $(mathbb Qtimesmathbb Q)/(mathbb Ztimes=)$ is not homeomorphic to $(mathbb Q/mathbb Z)times(mathbb Q/=)$?Quotient spaces homeomorphic to the realsProve that $(Xtimes [0,1])/sim'$ is homeomorphic to $(X/sim)times[0,1].$RP2 is homeomorphic to $(I times I)/ sim$Homeomorphism between a quotient topology on $mathbbR^2$ and $mathbbR$













0












$begingroup$



This question already has an answer here:



  • $(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$

    1 answer



Let $X=S^1times I$ where $S^1subset mathbbC$ and $I=[0,1]subset mathbbR$ also define the equivalence relation ~ by $(x,t)$~$(y,s)$ if and only if $xt=ys$.



Prove that $X$/~ is homeomorphic to the unit disc $D^2=xin mathbbR^2: =x$ with the induced topology.



Any hints would be appreciated.










share|cite|improve this question











$endgroup$



marked as duplicate by Paul Frost, Jyrki Lahtonen, mrtaurho, Song, Ramiro 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • 1




    $begingroup$
    what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
    $endgroup$
    – William
    Mar 9 at 21:12










  • $begingroup$
    Sorry $Isubset mathbbR$ i will fix it
    $endgroup$
    – Alfdav
    Mar 9 at 21:21






  • 2




    $begingroup$
    Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
    $endgroup$
    – Connor Malin
    Mar 9 at 21:38















0












$begingroup$



This question already has an answer here:



  • $(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$

    1 answer



Let $X=S^1times I$ where $S^1subset mathbbC$ and $I=[0,1]subset mathbbR$ also define the equivalence relation ~ by $(x,t)$~$(y,s)$ if and only if $xt=ys$.



Prove that $X$/~ is homeomorphic to the unit disc $D^2=xin mathbbR^2: =x$ with the induced topology.



Any hints would be appreciated.










share|cite|improve this question











$endgroup$



marked as duplicate by Paul Frost, Jyrki Lahtonen, mrtaurho, Song, Ramiro 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • 1




    $begingroup$
    what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
    $endgroup$
    – William
    Mar 9 at 21:12










  • $begingroup$
    Sorry $Isubset mathbbR$ i will fix it
    $endgroup$
    – Alfdav
    Mar 9 at 21:21






  • 2




    $begingroup$
    Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
    $endgroup$
    – Connor Malin
    Mar 9 at 21:38













0












0








0





$begingroup$



This question already has an answer here:



  • $(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$

    1 answer



Let $X=S^1times I$ where $S^1subset mathbbC$ and $I=[0,1]subset mathbbR$ also define the equivalence relation ~ by $(x,t)$~$(y,s)$ if and only if $xt=ys$.



Prove that $X$/~ is homeomorphic to the unit disc $D^2=xin mathbbR^2: =x$ with the induced topology.



Any hints would be appreciated.










share|cite|improve this question











$endgroup$





This question already has an answer here:



  • $(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$

    1 answer



Let $X=S^1times I$ where $S^1subset mathbbC$ and $I=[0,1]subset mathbbR$ also define the equivalence relation ~ by $(x,t)$~$(y,s)$ if and only if $xt=ys$.



Prove that $X$/~ is homeomorphic to the unit disc $D^2=xin mathbbR^2: =x$ with the induced topology.



Any hints would be appreciated.





This question already has an answer here:



  • $(S^1 times [0,1])/sim$ homeomorphic to unit disk $ D^2$

    1 answer







general-topology algebraic-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Paul Frost

11.6k3934




11.6k3934










asked Mar 9 at 21:06









AlfdavAlfdav

1027




1027




marked as duplicate by Paul Frost, Jyrki Lahtonen, mrtaurho, Song, Ramiro 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Paul Frost, Jyrki Lahtonen, mrtaurho, Song, Ramiro 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









  • 1




    $begingroup$
    what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
    $endgroup$
    – William
    Mar 9 at 21:12










  • $begingroup$
    Sorry $Isubset mathbbR$ i will fix it
    $endgroup$
    – Alfdav
    Mar 9 at 21:21






  • 2




    $begingroup$
    Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
    $endgroup$
    – Connor Malin
    Mar 9 at 21:38












  • 1




    $begingroup$
    what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
    $endgroup$
    – William
    Mar 9 at 21:12










  • $begingroup$
    Sorry $Isubset mathbbR$ i will fix it
    $endgroup$
    – Alfdav
    Mar 9 at 21:21






  • 2




    $begingroup$
    Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
    $endgroup$
    – Connor Malin
    Mar 9 at 21:38







1




1




$begingroup$
what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
$endgroup$
– William
Mar 9 at 21:12




$begingroup$
what do you mean by $xt = ys$? Is $I$ also being considered a subset of $mathbbC$?
$endgroup$
– William
Mar 9 at 21:12












$begingroup$
Sorry $Isubset mathbbR$ i will fix it
$endgroup$
– Alfdav
Mar 9 at 21:21




$begingroup$
Sorry $Isubset mathbbR$ i will fix it
$endgroup$
– Alfdav
Mar 9 at 21:21




2




2




$begingroup$
Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
$endgroup$
– Connor Malin
Mar 9 at 21:38




$begingroup$
Try imagining the equivalence relation on X. It identifies less points then you might expect because if $x neq y$ on the unit circle the only scalars you can multiply $x,y$ by to get them equal are 0 and 0. This construction is called the cone of $S^1$ which unsurprisingly is homeomorphic to a cone. So really what you want to show is that a cone is homeomorphic to a disc.
$endgroup$
– Connor Malin
Mar 9 at 21:38










1 Answer
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$begingroup$

Consider the function $fcolon S^1 times I to D^2$ defined by $f(x, t) = xt$. Show $f(x, t) = f(y, s)$ iff the pairs are equivalent, so $f$ induces a function on the quotient space



$$barfcolon (S^1 times I)/sim to D^2 $$



Now show that this function is a homeomorphism.






share|cite|improve this answer











$endgroup$



















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Consider the function $fcolon S^1 times I to D^2$ defined by $f(x, t) = xt$. Show $f(x, t) = f(y, s)$ iff the pairs are equivalent, so $f$ induces a function on the quotient space



    $$barfcolon (S^1 times I)/sim to D^2 $$



    Now show that this function is a homeomorphism.






    share|cite|improve this answer











    $endgroup$

















      3












      $begingroup$

      Consider the function $fcolon S^1 times I to D^2$ defined by $f(x, t) = xt$. Show $f(x, t) = f(y, s)$ iff the pairs are equivalent, so $f$ induces a function on the quotient space



      $$barfcolon (S^1 times I)/sim to D^2 $$



      Now show that this function is a homeomorphism.






      share|cite|improve this answer











      $endgroup$















        3












        3








        3





        $begingroup$

        Consider the function $fcolon S^1 times I to D^2$ defined by $f(x, t) = xt$. Show $f(x, t) = f(y, s)$ iff the pairs are equivalent, so $f$ induces a function on the quotient space



        $$barfcolon (S^1 times I)/sim to D^2 $$



        Now show that this function is a homeomorphism.






        share|cite|improve this answer











        $endgroup$



        Consider the function $fcolon S^1 times I to D^2$ defined by $f(x, t) = xt$. Show $f(x, t) = f(y, s)$ iff the pairs are equivalent, so $f$ induces a function on the quotient space



        $$barfcolon (S^1 times I)/sim to D^2 $$



        Now show that this function is a homeomorphism.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 9 at 21:52

























        answered Mar 9 at 21:45









        WilliamWilliam

        2,6051224




        2,6051224













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