How to show that $p(x)=(cos2pi x,sin2pi x)$ is a homeomorphism?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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Let $p:(0,1)to S^1-(0,1)$
Defined by
$p(x)=(cos2pi x,sin2pi x)$



How to show that this map is a homeomorphism?
Note that by $S^1$ I mean a unit circle in $R^2$ and $(0,1)$ it's North pole.










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  • Do you mean $S^1-(1,0)$?
    – Lord Shark the Unknown
    Aug 30 at 6:22










  • It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
    – Uncool
    Aug 30 at 6:24










  • No, it isn't. What is $p(1/4)$ according to your definition?
    – Lord Shark the Unknown
    Aug 30 at 6:25










  • Oops! Yes I mean (1,0)
    – Uncool
    Aug 30 at 6:30














up vote
0
down vote

favorite
1












Let $p:(0,1)to S^1-(0,1)$
Defined by
$p(x)=(cos2pi x,sin2pi x)$



How to show that this map is a homeomorphism?
Note that by $S^1$ I mean a unit circle in $R^2$ and $(0,1)$ it's North pole.










share|cite|improve this question























  • Do you mean $S^1-(1,0)$?
    – Lord Shark the Unknown
    Aug 30 at 6:22










  • It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
    – Uncool
    Aug 30 at 6:24










  • No, it isn't. What is $p(1/4)$ according to your definition?
    – Lord Shark the Unknown
    Aug 30 at 6:25










  • Oops! Yes I mean (1,0)
    – Uncool
    Aug 30 at 6:30












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





Let $p:(0,1)to S^1-(0,1)$
Defined by
$p(x)=(cos2pi x,sin2pi x)$



How to show that this map is a homeomorphism?
Note that by $S^1$ I mean a unit circle in $R^2$ and $(0,1)$ it's North pole.










share|cite|improve this question















Let $p:(0,1)to S^1-(0,1)$
Defined by
$p(x)=(cos2pi x,sin2pi x)$



How to show that this map is a homeomorphism?
Note that by $S^1$ I mean a unit circle in $R^2$ and $(0,1)$ it's North pole.







general-topology






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edited Aug 30 at 6:22









Lord Shark the Unknown

88.8k955115




88.8k955115










asked Aug 30 at 6:16









Uncool

7417




7417











  • Do you mean $S^1-(1,0)$?
    – Lord Shark the Unknown
    Aug 30 at 6:22










  • It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
    – Uncool
    Aug 30 at 6:24










  • No, it isn't. What is $p(1/4)$ according to your definition?
    – Lord Shark the Unknown
    Aug 30 at 6:25










  • Oops! Yes I mean (1,0)
    – Uncool
    Aug 30 at 6:30
















  • Do you mean $S^1-(1,0)$?
    – Lord Shark the Unknown
    Aug 30 at 6:22










  • It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
    – Uncool
    Aug 30 at 6:24










  • No, it isn't. What is $p(1/4)$ according to your definition?
    – Lord Shark the Unknown
    Aug 30 at 6:25










  • Oops! Yes I mean (1,0)
    – Uncool
    Aug 30 at 6:30















Do you mean $S^1-(1,0)$?
– Lord Shark the Unknown
Aug 30 at 6:22




Do you mean $S^1-(1,0)$?
– Lord Shark the Unknown
Aug 30 at 6:22












It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
– Uncool
Aug 30 at 6:24




It's the same thing I want to show that a circle minus one point is homeomorphic to an open interval
– Uncool
Aug 30 at 6:24












No, it isn't. What is $p(1/4)$ according to your definition?
– Lord Shark the Unknown
Aug 30 at 6:25




No, it isn't. What is $p(1/4)$ according to your definition?
– Lord Shark the Unknown
Aug 30 at 6:25












Oops! Yes I mean (1,0)
– Uncool
Aug 30 at 6:30




Oops! Yes I mean (1,0)
– Uncool
Aug 30 at 6:30










3 Answers
3






active

oldest

votes

















up vote
2
down vote













Use $p(x)=e^2pi i x$. Note that $p(0)=p(1)=1$.






share|cite|improve this answer




















  • Not sure why you give me this. I want the above function
    – Uncool
    Aug 30 at 6:25










  • It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
    – Amrat A
    Aug 30 at 6:27











  • Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
    – Amrat A
    Aug 30 at 6:32











  • Oa.. ok I'll try
    – Uncool
    Aug 30 at 6:32

















up vote
1
down vote













We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.



1) Note that for $x,y in (0,1)$ $$P(x)=P(y) implies (cos ( 2pi x, sin ( 2pi x) )= (cos ( 2pi y, sin ( 2pi y) ) implies x=y $$
$
2) If $$(cos ( 2pi x), sin ( 2pi x) ) in S^1-(0,1)$ then $0<x<1$



3) Both function $P$ and its inverse function are continuous.



Thus $P$ it is a homeomorphism.






share|cite|improve this answer



























    up vote
    0
    down vote













    It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $sqrt(cos(2pi y)-cos(2pi x))^2+(sin(2pi y)-sin(2pi x))^2=sqrt2(1-cos(2pi x-2pi y)$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0Leftrightarrow x=y$ and $|x-y|<epsilonLeftrightarrow d(p(x),p(y))<sqrt2(1-cos(2piepsilon)$ We know that $p$ is a diffeomorphism.






    share|cite|improve this answer




















    • What about the inverse? How to show that it's continuous and before that how to define it.
      – Uncool
      Aug 30 at 7:17











    • In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
      – Hugo
      Aug 30 at 8:22










    • Since it is equivalent, we can from left to the right ,and right to the left.
      – Hugo
      Aug 30 at 8:27










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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote













    Use $p(x)=e^2pi i x$. Note that $p(0)=p(1)=1$.






    share|cite|improve this answer




















    • Not sure why you give me this. I want the above function
      – Uncool
      Aug 30 at 6:25










    • It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
      – Amrat A
      Aug 30 at 6:27











    • Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
      – Amrat A
      Aug 30 at 6:32











    • Oa.. ok I'll try
      – Uncool
      Aug 30 at 6:32














    up vote
    2
    down vote













    Use $p(x)=e^2pi i x$. Note that $p(0)=p(1)=1$.






    share|cite|improve this answer




















    • Not sure why you give me this. I want the above function
      – Uncool
      Aug 30 at 6:25










    • It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
      – Amrat A
      Aug 30 at 6:27











    • Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
      – Amrat A
      Aug 30 at 6:32











    • Oa.. ok I'll try
      – Uncool
      Aug 30 at 6:32












    up vote
    2
    down vote










    up vote
    2
    down vote









    Use $p(x)=e^2pi i x$. Note that $p(0)=p(1)=1$.






    share|cite|improve this answer












    Use $p(x)=e^2pi i x$. Note that $p(0)=p(1)=1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 30 at 6:21









    Amrat A

    1115




    1115











    • Not sure why you give me this. I want the above function
      – Uncool
      Aug 30 at 6:25










    • It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
      – Amrat A
      Aug 30 at 6:27











    • Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
      – Amrat A
      Aug 30 at 6:32











    • Oa.. ok I'll try
      – Uncool
      Aug 30 at 6:32
















    • Not sure why you give me this. I want the above function
      – Uncool
      Aug 30 at 6:25










    • It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
      – Amrat A
      Aug 30 at 6:27











    • Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
      – Amrat A
      Aug 30 at 6:32











    • Oa.. ok I'll try
      – Uncool
      Aug 30 at 6:32















    Not sure why you give me this. I want the above function
    – Uncool
    Aug 30 at 6:25




    Not sure why you give me this. I want the above function
    – Uncool
    Aug 30 at 6:25












    It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
    – Amrat A
    Aug 30 at 6:27





    It is the same. By Eular's formula : $e^ix=cos x+i sin x$ or in $mathbb R^2$ it is equal to $(cos x, sin x)$
    – Amrat A
    Aug 30 at 6:27













    Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
    – Amrat A
    Aug 30 at 6:32





    Anyways, if you like to use $p(x)=(cos2pi x,sin2pi x)$, then note that $cos^22pi x+ sin^22pi x=1$ so it satisfies the circle equation.. and it is periodic. What is the period?
    – Amrat A
    Aug 30 at 6:32













    Oa.. ok I'll try
    – Uncool
    Aug 30 at 6:32




    Oa.. ok I'll try
    – Uncool
    Aug 30 at 6:32










    up vote
    1
    down vote













    We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.



    1) Note that for $x,y in (0,1)$ $$P(x)=P(y) implies (cos ( 2pi x, sin ( 2pi x) )= (cos ( 2pi y, sin ( 2pi y) ) implies x=y $$
    $
    2) If $$(cos ( 2pi x), sin ( 2pi x) ) in S^1-(0,1)$ then $0<x<1$



    3) Both function $P$ and its inverse function are continuous.



    Thus $P$ it is a homeomorphism.






    share|cite|improve this answer
























      up vote
      1
      down vote













      We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.



      1) Note that for $x,y in (0,1)$ $$P(x)=P(y) implies (cos ( 2pi x, sin ( 2pi x) )= (cos ( 2pi y, sin ( 2pi y) ) implies x=y $$
      $
      2) If $$(cos ( 2pi x), sin ( 2pi x) ) in S^1-(0,1)$ then $0<x<1$



      3) Both function $P$ and its inverse function are continuous.



      Thus $P$ it is a homeomorphism.






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.



        1) Note that for $x,y in (0,1)$ $$P(x)=P(y) implies (cos ( 2pi x, sin ( 2pi x) )= (cos ( 2pi y, sin ( 2pi y) ) implies x=y $$
        $
        2) If $$(cos ( 2pi x), sin ( 2pi x) ) in S^1-(0,1)$ then $0<x<1$



        3) Both function $P$ and its inverse function are continuous.



        Thus $P$ it is a homeomorphism.






        share|cite|improve this answer












        We need to show that the function $P$ is one-to-one, onto, continuous with continuous inverse.



        1) Note that for $x,y in (0,1)$ $$P(x)=P(y) implies (cos ( 2pi x, sin ( 2pi x) )= (cos ( 2pi y, sin ( 2pi y) ) implies x=y $$
        $
        2) If $$(cos ( 2pi x), sin ( 2pi x) ) in S^1-(0,1)$ then $0<x<1$



        3) Both function $P$ and its inverse function are continuous.



        Thus $P$ it is a homeomorphism.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 30 at 7:29









        Mohammad Riazi-Kermani

        31k41853




        31k41853




















            up vote
            0
            down vote













            It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $sqrt(cos(2pi y)-cos(2pi x))^2+(sin(2pi y)-sin(2pi x))^2=sqrt2(1-cos(2pi x-2pi y)$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0Leftrightarrow x=y$ and $|x-y|<epsilonLeftrightarrow d(p(x),p(y))<sqrt2(1-cos(2piepsilon)$ We know that $p$ is a diffeomorphism.






            share|cite|improve this answer




















            • What about the inverse? How to show that it's continuous and before that how to define it.
              – Uncool
              Aug 30 at 7:17











            • In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
              – Hugo
              Aug 30 at 8:22










            • Since it is equivalent, we can from left to the right ,and right to the left.
              – Hugo
              Aug 30 at 8:27














            up vote
            0
            down vote













            It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $sqrt(cos(2pi y)-cos(2pi x))^2+(sin(2pi y)-sin(2pi x))^2=sqrt2(1-cos(2pi x-2pi y)$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0Leftrightarrow x=y$ and $|x-y|<epsilonLeftrightarrow d(p(x),p(y))<sqrt2(1-cos(2piepsilon)$ We know that $p$ is a diffeomorphism.






            share|cite|improve this answer




















            • What about the inverse? How to show that it's continuous and before that how to define it.
              – Uncool
              Aug 30 at 7:17











            • In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
              – Hugo
              Aug 30 at 8:22










            • Since it is equivalent, we can from left to the right ,and right to the left.
              – Hugo
              Aug 30 at 8:27












            up vote
            0
            down vote










            up vote
            0
            down vote









            It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $sqrt(cos(2pi y)-cos(2pi x))^2+(sin(2pi y)-sin(2pi x))^2=sqrt2(1-cos(2pi x-2pi y)$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0Leftrightarrow x=y$ and $|x-y|<epsilonLeftrightarrow d(p(x),p(y))<sqrt2(1-cos(2piepsilon)$ We know that $p$ is a diffeomorphism.






            share|cite|improve this answer












            It is obvious that $p$ is surjective. The distance of $p(x)$ and $p(y)$ is $sqrt(cos(2pi y)-cos(2pi x))^2+(sin(2pi y)-sin(2pi x))^2=sqrt2(1-cos(2pi x-2pi y)$. Note that $-1<x-y<1$ , from $d(p(x),p(y))=0Leftrightarrow x=y$ and $|x-y|<epsilonLeftrightarrow d(p(x),p(y))<sqrt2(1-cos(2piepsilon)$ We know that $p$ is a diffeomorphism.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 30 at 6:40









            Hugo

            1029




            1029











            • What about the inverse? How to show that it's continuous and before that how to define it.
              – Uncool
              Aug 30 at 7:17











            • In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
              – Hugo
              Aug 30 at 8:22










            • Since it is equivalent, we can from left to the right ,and right to the left.
              – Hugo
              Aug 30 at 8:27
















            • What about the inverse? How to show that it's continuous and before that how to define it.
              – Uncool
              Aug 30 at 7:17











            • In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
              – Hugo
              Aug 30 at 8:22










            • Since it is equivalent, we can from left to the right ,and right to the left.
              – Hugo
              Aug 30 at 8:27















            What about the inverse? How to show that it's continuous and before that how to define it.
            – Uncool
            Aug 30 at 7:17





            What about the inverse? How to show that it's continuous and before that how to define it.
            – Uncool
            Aug 30 at 7:17













            In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
            – Hugo
            Aug 30 at 8:22




            In deed we don't need to write the formula of the converse. The first equivalent symble implies that it is invertible. The second symble implies both of them are continuous.
            – Hugo
            Aug 30 at 8:22












            Since it is equivalent, we can from left to the right ,and right to the left.
            – Hugo
            Aug 30 at 8:27




            Since it is equivalent, we can from left to the right ,and right to the left.
            – Hugo
            Aug 30 at 8:27

















             

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