Show that the quotient space is homeomorphic to the n-disc












4












$begingroup$



Let $sim$ denote the equivalence relation on the $n$-sphere $S^n$ defined via
$$
(x_1,dots,x_n,x_{n+1})sim(x_1,dots,x_n,−x_{n+1}):text{ for all }: (x_1,dots,x_{n+1})in S^n.
$$



Show that the quotient space $S^n/sim$ is homeomorphic to the $n$-disc $D^n$.




I know that Ii have to show that there exists a bijective function $f: (S^n/sim) to D^n$ that is continuous and that the pre-image $f^{-1}$ is continuous as well, but I can't advance from here, any tips?






general-topology







share|cite|improve this question











$endgroup$

















    4












    $begingroup$



    Let $sim$ denote the equivalence relation on the $n$-sphere $S^n$ defined via
    $$
    (x_1,dots,x_n,x_{n+1})sim(x_1,dots,x_n,−x_{n+1}):text{ for all }: (x_1,dots,x_{n+1})in S^n.
    $$



    Show that the quotient space $S^n/sim$ is homeomorphic to the $n$-disc $D^n$.




    I know that Ii have to show that there exists a bijective function $f: (S^n/sim) to D^n$ that is continuous and that the pre-image $f^{-1}$ is continuous as well, but I can't advance from here, any tips?






    general-topology







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$



      Let $sim$ denote the equivalence relation on the $n$-sphere $S^n$ defined via
      $$
      (x_1,dots,x_n,x_{n+1})sim(x_1,dots,x_n,−x_{n+1}):text{ for all }: (x_1,dots,x_{n+1})in S^n.
      $$



      Show that the quotient space $S^n/sim$ is homeomorphic to the $n$-disc $D^n$.




      I know that Ii have to show that there exists a bijective function $f: (S^n/sim) to D^n$ that is continuous and that the pre-image $f^{-1}$ is continuous as well, but I can't advance from here, any tips?






      general-topology







      share|cite|improve this question











      $endgroup$





      Let $sim$ denote the equivalence relation on the $n$-sphere $S^n$ defined via
      $$
      (x_1,dots,x_n,x_{n+1})sim(x_1,dots,x_n,−x_{n+1}):text{ for all }: (x_1,dots,x_{n+1})in S^n.
      $$



      Show that the quotient space $S^n/sim$ is homeomorphic to the $n$-disc $D^n$.




      I know that Ii have to show that there exists a bijective function $f: (S^n/sim) to D^n$ that is continuous and that the pre-image $f^{-1}$ is continuous as well, but I can't advance from here, any tips?






      general-topology




      general-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 13:10









      Glorfindel

      3,41981830




      3,41981830










      asked Dec 11 '18 at 12:55









      Math MatichMath Matich

      322




      322






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Hint: Define$$begin{array}{rccc}fcolon&S^n/sim&longrightarrow&D^n\&bigl[(x_1,ldots,x_n,x_{n+1})bigr]&mapsto&(x_1,ldots,x_n).end{array}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay, i see, but my struggle is to advance from here, how do i show that the function is continuous and furthermore that the pre image is?
            $endgroup$
            – Math Matich
            Dec 11 '18 at 13:40










          • $begingroup$
            Two things. First of all: it is not pre-image, it's the inverse function of $f$. Secondly: you asked for a tip and I provided one.
            $endgroup$
            – José Carlos Santos
            Dec 11 '18 at 13:44










          • $begingroup$
            Okay, thanks. I'll try.
            $endgroup$
            – Math Matich
            Dec 11 '18 at 14:10










          • $begingroup$
            @MathMatich the inverse function is clear: map $(x_1, x_2,ldots, x_n)$ to the class of $(x_1,x_2,ldots,x_n, 1-sqrt{sum_{i=1}^n x^2_i})$. $f$ being continuous is clear (it's a projection in essence) and compactness does the rest as soon as you know it's a bijection.
            $endgroup$
            – Henno Brandsma
            Dec 11 '18 at 15:42












          • $begingroup$
            I suppose that it's a typo and that it should be$$left(x_1,x_2,ldots,x_n,sqrt{1-sum_{n=1}^n{x_n}^2}right).$$
            $endgroup$
            – José Carlos Santos
            Dec 12 '18 at 18:04













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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          3












          $begingroup$

          Hint: Define$$begin{array}{rccc}fcolon&S^n/sim&longrightarrow&D^n\&bigl[(x_1,ldots,x_n,x_{n+1})bigr]&mapsto&(x_1,ldots,x_n).end{array}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay, i see, but my struggle is to advance from here, how do i show that the function is continuous and furthermore that the pre image is?
            $endgroup$
            – Math Matich
            Dec 11 '18 at 13:40










          • $begingroup$
            Two things. First of all: it is not pre-image, it's the inverse function of $f$. Secondly: you asked for a tip and I provided one.
            $endgroup$
            – José Carlos Santos
            Dec 11 '18 at 13:44










          • $begingroup$
            Okay, thanks. I'll try.
            $endgroup$
            – Math Matich
            Dec 11 '18 at 14:10










          • $begingroup$
            @MathMatich the inverse function is clear: map $(x_1, x_2,ldots, x_n)$ to the class of $(x_1,x_2,ldots,x_n, 1-sqrt{sum_{i=1}^n x^2_i})$. $f$ being continuous is clear (it's a projection in essence) and compactness does the rest as soon as you know it's a bijection.
            $endgroup$
            – Henno Brandsma
            Dec 11 '18 at 15:42












          • $begingroup$
            I suppose that it's a typo and that it should be$$left(x_1,x_2,ldots,x_n,sqrt{1-sum_{n=1}^n{x_n}^2}right).$$
            $endgroup$
            – José Carlos Santos
            Dec 12 '18 at 18:04


















          3












          $begingroup$

          Hint: Define$$begin{array}{rccc}fcolon&S^n/sim&longrightarrow&D^n\&bigl[(x_1,ldots,x_n,x_{n+1})bigr]&mapsto&(x_1,ldots,x_n).end{array}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay, i see, but my struggle is to advance from here, how do i show that the function is continuous and furthermore that the pre image is?
            $endgroup$
            – Math Matich
            Dec 11 '18 at 13:40










          • $begingroup$
            Two things. First of all: it is not pre-image, it's the inverse function of $f$. Secondly: you asked for a tip and I provided one.
            $endgroup$
            – José Carlos Santos
            Dec 11 '18 at 13:44










          • $begingroup$
            Okay, thanks. I'll try.
            $endgroup$
            – Math Matich
            Dec 11 '18 at 14:10










          • $begingroup$
            @MathMatich the inverse function is clear: map $(x_1, x_2,ldots, x_n)$ to the class of $(x_1,x_2,ldots,x_n, 1-sqrt{sum_{i=1}^n x^2_i})$. $f$ being continuous is clear (it's a projection in essence) and compactness does the rest as soon as you know it's a bijection.
            $endgroup$
            – Henno Brandsma
            Dec 11 '18 at 15:42












          • $begingroup$
            I suppose that it's a typo and that it should be$$left(x_1,x_2,ldots,x_n,sqrt{1-sum_{n=1}^n{x_n}^2}right).$$
            $endgroup$
            – José Carlos Santos
            Dec 12 '18 at 18:04
















          3












          3








          3





          $begingroup$

          Hint: Define$$begin{array}{rccc}fcolon&S^n/sim&longrightarrow&D^n\&bigl[(x_1,ldots,x_n,x_{n+1})bigr]&mapsto&(x_1,ldots,x_n).end{array}$$






          share|cite|improve this answer









          $endgroup$



          Hint: Define$$begin{array}{rccc}fcolon&S^n/sim&longrightarrow&D^n\&bigl[(x_1,ldots,x_n,x_{n+1})bigr]&mapsto&(x_1,ldots,x_n).end{array}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 13:03









          José Carlos SantosJosé Carlos Santos

          162k22130233




          162k22130233












          • $begingroup$
            Okay, i see, but my struggle is to advance from here, how do i show that the function is continuous and furthermore that the pre image is?
            $endgroup$
            – Math Matich
            Dec 11 '18 at 13:40










          • $begingroup$
            Two things. First of all: it is not pre-image, it's the inverse function of $f$. Secondly: you asked for a tip and I provided one.
            $endgroup$
            – José Carlos Santos
            Dec 11 '18 at 13:44










          • $begingroup$
            Okay, thanks. I'll try.
            $endgroup$
            – Math Matich
            Dec 11 '18 at 14:10










          • $begingroup$
            @MathMatich the inverse function is clear: map $(x_1, x_2,ldots, x_n)$ to the class of $(x_1,x_2,ldots,x_n, 1-sqrt{sum_{i=1}^n x^2_i})$. $f$ being continuous is clear (it's a projection in essence) and compactness does the rest as soon as you know it's a bijection.
            $endgroup$
            – Henno Brandsma
            Dec 11 '18 at 15:42












          • $begingroup$
            I suppose that it's a typo and that it should be$$left(x_1,x_2,ldots,x_n,sqrt{1-sum_{n=1}^n{x_n}^2}right).$$
            $endgroup$
            – José Carlos Santos
            Dec 12 '18 at 18:04




















          • $begingroup$
            Okay, i see, but my struggle is to advance from here, how do i show that the function is continuous and furthermore that the pre image is?
            $endgroup$
            – Math Matich
            Dec 11 '18 at 13:40










          • $begingroup$
            Two things. First of all: it is not pre-image, it's the inverse function of $f$. Secondly: you asked for a tip and I provided one.
            $endgroup$
            – José Carlos Santos
            Dec 11 '18 at 13:44










          • $begingroup$
            Okay, thanks. I'll try.
            $endgroup$
            – Math Matich
            Dec 11 '18 at 14:10










          • $begingroup$
            @MathMatich the inverse function is clear: map $(x_1, x_2,ldots, x_n)$ to the class of $(x_1,x_2,ldots,x_n, 1-sqrt{sum_{i=1}^n x^2_i})$. $f$ being continuous is clear (it's a projection in essence) and compactness does the rest as soon as you know it's a bijection.
            $endgroup$
            – Henno Brandsma
            Dec 11 '18 at 15:42












          • $begingroup$
            I suppose that it's a typo and that it should be$$left(x_1,x_2,ldots,x_n,sqrt{1-sum_{n=1}^n{x_n}^2}right).$$
            $endgroup$
            – José Carlos Santos
            Dec 12 '18 at 18:04


















          $begingroup$
          Okay, i see, but my struggle is to advance from here, how do i show that the function is continuous and furthermore that the pre image is?
          $endgroup$
          – Math Matich
          Dec 11 '18 at 13:40




          $begingroup$
          Okay, i see, but my struggle is to advance from here, how do i show that the function is continuous and furthermore that the pre image is?
          $endgroup$
          – Math Matich
          Dec 11 '18 at 13:40












          $begingroup$
          Two things. First of all: it is not pre-image, it's the inverse function of $f$. Secondly: you asked for a tip and I provided one.
          $endgroup$
          – José Carlos Santos
          Dec 11 '18 at 13:44




          $begingroup$
          Two things. First of all: it is not pre-image, it's the inverse function of $f$. Secondly: you asked for a tip and I provided one.
          $endgroup$
          – José Carlos Santos
          Dec 11 '18 at 13:44












          $begingroup$
          Okay, thanks. I'll try.
          $endgroup$
          – Math Matich
          Dec 11 '18 at 14:10




          $begingroup$
          Okay, thanks. I'll try.
          $endgroup$
          – Math Matich
          Dec 11 '18 at 14:10












          $begingroup$
          @MathMatich the inverse function is clear: map $(x_1, x_2,ldots, x_n)$ to the class of $(x_1,x_2,ldots,x_n, 1-sqrt{sum_{i=1}^n x^2_i})$. $f$ being continuous is clear (it's a projection in essence) and compactness does the rest as soon as you know it's a bijection.
          $endgroup$
          – Henno Brandsma
          Dec 11 '18 at 15:42






          $begingroup$
          @MathMatich the inverse function is clear: map $(x_1, x_2,ldots, x_n)$ to the class of $(x_1,x_2,ldots,x_n, 1-sqrt{sum_{i=1}^n x^2_i})$. $f$ being continuous is clear (it's a projection in essence) and compactness does the rest as soon as you know it's a bijection.
          $endgroup$
          – Henno Brandsma
          Dec 11 '18 at 15:42














          $begingroup$
          I suppose that it's a typo and that it should be$$left(x_1,x_2,ldots,x_n,sqrt{1-sum_{n=1}^n{x_n}^2}right).$$
          $endgroup$
          – José Carlos Santos
          Dec 12 '18 at 18:04






          $begingroup$
          I suppose that it's a typo and that it should be$$left(x_1,x_2,ldots,x_n,sqrt{1-sum_{n=1}^n{x_n}^2}right).$$
          $endgroup$
          – José Carlos Santos
          Dec 12 '18 at 18:04




















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