How to understand this example in Do Carmo?











up vote
5
down vote

favorite












I'm reading the book $Riemannian$ $Geometry$ written by Do Carmo. Here is an example that I cannot understand the explanation he gave.
enter image description here



enter image description here



enter image description here



I really don't understand what he said about why $alpha$ is not an embedding...
No worry about my knowledge on topology, can anyone help me ''translate'' it to the common language that's easy to understand?






general-topology differential-geometry riemannian-geometry curves







share|cite|improve this question




























    up vote
    5
    down vote

    favorite












    I'm reading the book $Riemannian$ $Geometry$ written by Do Carmo. Here is an example that I cannot understand the explanation he gave.
    enter image description here



    enter image description here



    enter image description here



    I really don't understand what he said about why $alpha$ is not an embedding...
    No worry about my knowledge on topology, can anyone help me ''translate'' it to the common language that's easy to understand?






    general-topology differential-geometry riemannian-geometry curves







    share|cite|improve this question


























      up vote
      5
      down vote

      favorite









      up vote
      5
      down vote

      favorite











      I'm reading the book $Riemannian$ $Geometry$ written by Do Carmo. Here is an example that I cannot understand the explanation he gave.
      enter image description here



      enter image description here



      enter image description here



      I really don't understand what he said about why $alpha$ is not an embedding...
      No worry about my knowledge on topology, can anyone help me ''translate'' it to the common language that's easy to understand?






      general-topology differential-geometry riemannian-geometry curves







      share|cite|improve this question















      I'm reading the book $Riemannian$ $Geometry$ written by Do Carmo. Here is an example that I cannot understand the explanation he gave.
      enter image description here



      enter image description here



      enter image description here



      I really don't understand what he said about why $alpha$ is not an embedding...
      No worry about my knowledge on topology, can anyone help me ''translate'' it to the common language that's easy to understand?






      general-topology differential-geometry riemannian-geometry curves




      general-topology differential-geometry riemannian-geometry curves






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday









      José Carlos Santos

      137k17110199




      137k17110199










      asked yesterday









      user450201

      648




      648






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          8
          down vote



          accepted










          The set $C=alphabigl((-3,0)bigr)$ has two topologies:




          • the topology it inherits from the usual topology in $mathbb{R}^2$: a set $Asubset C$ is open if there is an open subset of $A^star$ of $mathbb{R}^2$ such that $A=A^starcap C$.

          • the topology in gets from $(-3,0)$: a set $Asubset C$ is open if there is an open subcet $A^star$ of $(-3,0)$ such that $A=alpha(A^star)$.


          Then, Do Carmo explains why these two topologies are distinct: the second one is locally connected, whereas the first one is no.






          share|cite|improve this answer



















          • 1




            Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
            – user450201
            yesterday










          • That would be correct. On the other hand, that is what Do Carmo is claiming.
            – José Carlos Santos
            yesterday











          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',
          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%2f2995192%2fhow-to-understand-this-example-in-do-carmo%23new-answer', 'question_page');
          }
          );

          Post as a guest


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          8
          down vote



          accepted










          The set $C=alphabigl((-3,0)bigr)$ has two topologies:




          • the topology it inherits from the usual topology in $mathbb{R}^2$: a set $Asubset C$ is open if there is an open subset of $A^star$ of $mathbb{R}^2$ such that $A=A^starcap C$.

          • the topology in gets from $(-3,0)$: a set $Asubset C$ is open if there is an open subcet $A^star$ of $(-3,0)$ such that $A=alpha(A^star)$.


          Then, Do Carmo explains why these two topologies are distinct: the second one is locally connected, whereas the first one is no.






          share|cite|improve this answer



















          • 1




            Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
            – user450201
            yesterday










          • That would be correct. On the other hand, that is what Do Carmo is claiming.
            – José Carlos Santos
            yesterday















          up vote
          8
          down vote



          accepted










          The set $C=alphabigl((-3,0)bigr)$ has two topologies:




          • the topology it inherits from the usual topology in $mathbb{R}^2$: a set $Asubset C$ is open if there is an open subset of $A^star$ of $mathbb{R}^2$ such that $A=A^starcap C$.

          • the topology in gets from $(-3,0)$: a set $Asubset C$ is open if there is an open subcet $A^star$ of $(-3,0)$ such that $A=alpha(A^star)$.


          Then, Do Carmo explains why these two topologies are distinct: the second one is locally connected, whereas the first one is no.






          share|cite|improve this answer



















          • 1




            Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
            – user450201
            yesterday










          • That would be correct. On the other hand, that is what Do Carmo is claiming.
            – José Carlos Santos
            yesterday













          up vote
          8
          down vote



          accepted







          up vote
          8
          down vote



          accepted






          The set $C=alphabigl((-3,0)bigr)$ has two topologies:




          • the topology it inherits from the usual topology in $mathbb{R}^2$: a set $Asubset C$ is open if there is an open subset of $A^star$ of $mathbb{R}^2$ such that $A=A^starcap C$.

          • the topology in gets from $(-3,0)$: a set $Asubset C$ is open if there is an open subcet $A^star$ of $(-3,0)$ such that $A=alpha(A^star)$.


          Then, Do Carmo explains why these two topologies are distinct: the second one is locally connected, whereas the first one is no.






          share|cite|improve this answer














          The set $C=alphabigl((-3,0)bigr)$ has two topologies:




          • the topology it inherits from the usual topology in $mathbb{R}^2$: a set $Asubset C$ is open if there is an open subset of $A^star$ of $mathbb{R}^2$ such that $A=A^starcap C$.

          • the topology in gets from $(-3,0)$: a set $Asubset C$ is open if there is an open subcet $A^star$ of $(-3,0)$ such that $A=alpha(A^star)$.


          Then, Do Carmo explains why these two topologies are distinct: the second one is locally connected, whereas the first one is no.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          José Carlos Santos

          137k17110199




          137k17110199








          • 1




            Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
            – user450201
            yesterday










          • That would be correct. On the other hand, that is what Do Carmo is claiming.
            – José Carlos Santos
            yesterday














          • 1




            Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
            – user450201
            yesterday










          • That would be correct. On the other hand, that is what Do Carmo is claiming.
            – José Carlos Santos
            yesterday








          1




          1




          Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
          – user450201
          yesterday




          Can I just say that (-3,0) is locally connected while the image of alpha is not, so alpha is not a homeomorphism?
          – user450201
          yesterday












          That would be correct. On the other hand, that is what Do Carmo is claiming.
          – José Carlos Santos
          yesterday




          That would be correct. On the other hand, that is what Do Carmo is claiming.
          – José Carlos Santos
          yesterday


















           

          draft saved


          draft discarded



















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2995192%2fhow-to-understand-this-example-in-do-carmo%23new-answer', 'question_page');
          }
          );

          Post as a guest






































































          -

          Popular posts from this blog

          Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

          Image
          2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
          Read more

          ComboBox Display Member on multiple fields

          Image
          2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
          Read more

          Is it possible to collect Nectar points via Trainline?

          Image
          2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
          Read more