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













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035259%2fshow-that-the-quotient-space-is-homeomorphic-to-the-n-disc%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          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


















          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




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3035259%2fshow-that-the-quotient-space-is-homeomorphic-to-the-n-disc%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Bundesstraße 106

          Image
          Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
          Read more

          Verónica Boquete

          Image
          Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
          Read more

          Ida-Boy-Ed-Garten

          Image
          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
          Read more