Quotient of $[-1,1]$ homeomorphic to unit circle

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Let $I:=[-1,1]$ and $Q$ the quotient space where we identified $-1$ and $1$. Show hat $Q$ is homeomorphic to the unit circle $S^1subseteq mathbbR^2$.



My first idea was to define $gcolon Ito S^1$ by $xmapsto (cos(2pi x), sin(2pi x))$, but then the map $f$ satisfying $g=fcirc p$, where $p$ is the quotient map, seems not to be injective. Any hints?




general-topology quotient-spaces




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    up vote
    2
    down vote

    favorite












    Let $I:=[-1,1]$ and $Q$ the quotient space where we identified $-1$ and $1$. Show hat $Q$ is homeomorphic to the unit circle $S^1subseteq mathbbR^2$.



    My first idea was to define $gcolon Ito S^1$ by $xmapsto (cos(2pi x), sin(2pi x))$, but then the map $f$ satisfying $g=fcirc p$, where $p$ is the quotient map, seems not to be injective. Any hints?




    general-topology quotient-spaces




    share|cite|improve this question





















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $I:=[-1,1]$ and $Q$ the quotient space where we identified $-1$ and $1$. Show hat $Q$ is homeomorphic to the unit circle $S^1subseteq mathbbR^2$.



      My first idea was to define $gcolon Ito S^1$ by $xmapsto (cos(2pi x), sin(2pi x))$, but then the map $f$ satisfying $g=fcirc p$, where $p$ is the quotient map, seems not to be injective. Any hints?




      general-topology quotient-spaces




      share|cite|improve this question











      Let $I:=[-1,1]$ and $Q$ the quotient space where we identified $-1$ and $1$. Show hat $Q$ is homeomorphic to the unit circle $S^1subseteq mathbbR^2$.



      My first idea was to define $gcolon Ito S^1$ by $xmapsto (cos(2pi x), sin(2pi x))$, but then the map $f$ satisfying $g=fcirc p$, where $p$ is the quotient map, seems not to be injective. Any hints?





      general-topology quotient-spaces





      share|cite|improve this question










      share|cite|improve this question




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      asked Jul 31 at 10:22









      mathstackuser

      1667




      1667




















          2 Answers
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          active

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          up vote
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          $$x mapsto (cos(pi x), sin(pi x))$$






          share|cite|improve this answer





















          • Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
            – mathstackuser
            Jul 31 at 10:55






          • 1




            Compact -> Hausdorff
            – Kenny Lau
            Jul 31 at 10:55










          • Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
            – mathstackuser
            Jul 31 at 11:18






          • 1




            "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
            – Kenny Lau
            Jul 31 at 11:19










          • Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
            – mathstackuser
            Jul 31 at 11:33

















          up vote
          1
          down vote













          I'm pretty sure you can show that $I=[-1,1] approx [0,1]$, say by $h : [-1,1] to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/sim$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X to Y$ which makes same identification as projection map $p : X to X/sim$. In your case define $q : [0,1] to BbbS^1$ by $s mapsto e^2pi is$.
          $requireAMScd$
          beginCD
          I @>h>> [0,1]\
          @VpVV @VVqV\
          I/sim @>>> BbbS^1
          endCD
          Since $q circ h$ is a quotient map that makes same identification as $p$, then $I=[-1,1] approx BbbS^1$ by Uniqueness of Quotient Topology.






          share|cite|improve this answer





















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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            $$x mapsto (cos(pi x), sin(pi x))$$






            share|cite|improve this answer





















            • Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
              – mathstackuser
              Jul 31 at 10:55






            • 1




              Compact -> Hausdorff
              – Kenny Lau
              Jul 31 at 10:55










            • Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
              – mathstackuser
              Jul 31 at 11:18






            • 1




              "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
              – Kenny Lau
              Jul 31 at 11:19










            • Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
              – mathstackuser
              Jul 31 at 11:33














            up vote
            3
            down vote



            accepted










            $$x mapsto (cos(pi x), sin(pi x))$$






            share|cite|improve this answer





















            • Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
              – mathstackuser
              Jul 31 at 10:55






            • 1




              Compact -> Hausdorff
              – Kenny Lau
              Jul 31 at 10:55










            • Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
              – mathstackuser
              Jul 31 at 11:18






            • 1




              "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
              – Kenny Lau
              Jul 31 at 11:19










            • Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
              – mathstackuser
              Jul 31 at 11:33












            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            $$x mapsto (cos(pi x), sin(pi x))$$






            share|cite|improve this answer













            $$x mapsto (cos(pi x), sin(pi x))$$







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 31 at 10:23









            Kenny Lau

            17.7k2156




            17.7k2156











            • Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
              – mathstackuser
              Jul 31 at 10:55






            • 1




              Compact -> Hausdorff
              – Kenny Lau
              Jul 31 at 10:55










            • Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
              – mathstackuser
              Jul 31 at 11:18






            • 1




              "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
              – Kenny Lau
              Jul 31 at 11:19










            • Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
              – mathstackuser
              Jul 31 at 11:33
















            • Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
              – mathstackuser
              Jul 31 at 10:55






            • 1




              Compact -> Hausdorff
              – Kenny Lau
              Jul 31 at 10:55










            • Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
              – mathstackuser
              Jul 31 at 11:18






            • 1




              "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
              – Kenny Lau
              Jul 31 at 11:19










            • Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
              – mathstackuser
              Jul 31 at 11:33















            Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
            – mathstackuser
            Jul 31 at 10:55




            Thanks! I can see that the induced map $f$ is continuous and bijective. How do I show that this is an open map?
            – mathstackuser
            Jul 31 at 10:55




            1




            1




            Compact -> Hausdorff
            – Kenny Lau
            Jul 31 at 10:55




            Compact -> Hausdorff
            – Kenny Lau
            Jul 31 at 10:55












            Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
            – mathstackuser
            Jul 31 at 11:18




            Can you please provide some more details? I am pretty new to topology... Which space is compact? And what do you mean with the arrow? Not all compact spaces are Hausdorf right? And how does this help?
            – mathstackuser
            Jul 31 at 11:18




            1




            1




            "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
            – Kenny Lau
            Jul 31 at 11:19




            "Compact -> Hausdorff" is the mnemonic for the theorem that if you have a continuous bijection from a compact space to a Hausdorff space, then it is a homeomorphism.
            – Kenny Lau
            Jul 31 at 11:19












            Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
            – mathstackuser
            Jul 31 at 11:33




            Thanks! And $Q$ is compact because it's the image of $I$ under the continuous function $p$ and $I$ is compact? And $S^1$ is Hausdorff because it's a subspace of $R^2$ which is Hausdorff?
            – mathstackuser
            Jul 31 at 11:33










            up vote
            1
            down vote













            I'm pretty sure you can show that $I=[-1,1] approx [0,1]$, say by $h : [-1,1] to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/sim$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X to Y$ which makes same identification as projection map $p : X to X/sim$. In your case define $q : [0,1] to BbbS^1$ by $s mapsto e^2pi is$.
            $requireAMScd$
            beginCD
            I @>h>> [0,1]\
            @VpVV @VVqV\
            I/sim @>>> BbbS^1
            endCD
            Since $q circ h$ is a quotient map that makes same identification as $p$, then $I=[-1,1] approx BbbS^1$ by Uniqueness of Quotient Topology.






            share|cite|improve this answer

























              up vote
              1
              down vote













              I'm pretty sure you can show that $I=[-1,1] approx [0,1]$, say by $h : [-1,1] to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/sim$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X to Y$ which makes same identification as projection map $p : X to X/sim$. In your case define $q : [0,1] to BbbS^1$ by $s mapsto e^2pi is$.
              $requireAMScd$
              beginCD
              I @>h>> [0,1]\
              @VpVV @VVqV\
              I/sim @>>> BbbS^1
              endCD
              Since $q circ h$ is a quotient map that makes same identification as $p$, then $I=[-1,1] approx BbbS^1$ by Uniqueness of Quotient Topology.






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                I'm pretty sure you can show that $I=[-1,1] approx [0,1]$, say by $h : [-1,1] to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/sim$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X to Y$ which makes same identification as projection map $p : X to X/sim$. In your case define $q : [0,1] to BbbS^1$ by $s mapsto e^2pi is$.
                $requireAMScd$
                beginCD
                I @>h>> [0,1]\
                @VpVV @VVqV\
                I/sim @>>> BbbS^1
                endCD
                Since $q circ h$ is a quotient map that makes same identification as $p$, then $I=[-1,1] approx BbbS^1$ by Uniqueness of Quotient Topology.






                share|cite|improve this answer













                I'm pretty sure you can show that $I=[-1,1] approx [0,1]$, say by $h : [-1,1] to [0,1]$ s.t. $h(-1)=0$ and $h(1)=1$. A common way to show that some quotient space $X/sim$ is homeomorphic to a top. space $Y$ is by exhibit a quotient map $q : X to Y$ which makes same identification as projection map $p : X to X/sim$. In your case define $q : [0,1] to BbbS^1$ by $s mapsto e^2pi is$.
                $requireAMScd$
                beginCD
                I @>h>> [0,1]\
                @VpVV @VVqV\
                I/sim @>>> BbbS^1
                endCD
                Since $q circ h$ is a quotient map that makes same identification as $p$, then $I=[-1,1] approx BbbS^1$ by Uniqueness of Quotient Topology.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 31 at 12:04









                Sou

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