Prove that $S^n /{p,q}$ is homotopy equivalent to $S^n vee S^1$.












2












$begingroup$



Prove that $S^n /{p, q}$ homotopy equivalent to $S^n ∨ S^1$




My attempt:



I can see the picture quite clearly: enter image description here



But how to write explicit homotopies ?



Thanks in Advance for help!






algebraic-topology homotopy-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
    $endgroup$
    – Neal
    Nov 3 '18 at 15:12










  • $begingroup$
    @Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
    $endgroup$
    – Cheerful Parsnip
    Nov 3 '18 at 16:37










  • $begingroup$
    @Neal Can you kindly post a full answer.
    $endgroup$
    – Coherent
    Nov 5 '18 at 4:40
















2












$begingroup$



Prove that $S^n /{p, q}$ homotopy equivalent to $S^n ∨ S^1$




My attempt:



I can see the picture quite clearly: enter image description here



But how to write explicit homotopies ?



Thanks in Advance for help!






algebraic-topology homotopy-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
    $endgroup$
    – Neal
    Nov 3 '18 at 15:12










  • $begingroup$
    @Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
    $endgroup$
    – Cheerful Parsnip
    Nov 3 '18 at 16:37










  • $begingroup$
    @Neal Can you kindly post a full answer.
    $endgroup$
    – Coherent
    Nov 5 '18 at 4:40














2












2








2


1



$begingroup$



Prove that $S^n /{p, q}$ homotopy equivalent to $S^n ∨ S^1$




My attempt:



I can see the picture quite clearly: enter image description here



But how to write explicit homotopies ?



Thanks in Advance for help!






algebraic-topology homotopy-theory







share|cite|improve this question











$endgroup$





Prove that $S^n /{p, q}$ homotopy equivalent to $S^n ∨ S^1$




My attempt:



I can see the picture quite clearly: enter image description here



But how to write explicit homotopies ?



Thanks in Advance for help!






algebraic-topology homotopy-theory




algebraic-topology homotopy-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 3 '18 at 15:10









Tyrone

4,50511225




4,50511225










asked Nov 3 '18 at 15:01









CoherentCoherent

1,230523




1,230523








  • 2




    $begingroup$
    You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
    $endgroup$
    – Neal
    Nov 3 '18 at 15:12










  • $begingroup$
    @Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
    $endgroup$
    – Cheerful Parsnip
    Nov 3 '18 at 16:37










  • $begingroup$
    @Neal Can you kindly post a full answer.
    $endgroup$
    – Coherent
    Nov 5 '18 at 4:40














  • 2




    $begingroup$
    You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
    $endgroup$
    – Neal
    Nov 3 '18 at 15:12










  • $begingroup$
    @Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
    $endgroup$
    – Cheerful Parsnip
    Nov 3 '18 at 16:37










  • $begingroup$
    @Neal Can you kindly post a full answer.
    $endgroup$
    – Coherent
    Nov 5 '18 at 4:40








2




2




$begingroup$
You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
$endgroup$
– Neal
Nov 3 '18 at 15:12




$begingroup$
You might argue that $S^n/{p,q}$ is homotopy equivalent to $S^ncup mbox{chord connecting $p,q$}$ (the latter deformation retracts to the former) and then argue that $S^n$ with a chord connecting $p,q$ is homotopy equivalent to $S^nwedge S^1$.
$endgroup$
– Neal
Nov 3 '18 at 15:12












$begingroup$
@Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
$endgroup$
– Cheerful Parsnip
Nov 3 '18 at 16:37




$begingroup$
@Jeroen: You are right that it's not homeomorphic to a wedge of a circle and sphere, but it is homotopy-equivalent.
$endgroup$
– Cheerful Parsnip
Nov 3 '18 at 16:37












$begingroup$
@Neal Can you kindly post a full answer.
$endgroup$
– Coherent
Nov 5 '18 at 4:40




$begingroup$
@Neal Can you kindly post a full answer.
$endgroup$
– Coherent
Nov 5 '18 at 4:40










2 Answers
2






active

oldest

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1












$begingroup$

We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $Xsimeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Ysimeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Ysimeq Y/B$, and $Y/B$ is homeomorphic to $S^nvee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $simeq$ you are done.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    For the inspiration, you should see Example 0.8 in Hatcher's book.



    Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 vee S^1$.



    You can generalize his argument using $CW$-complex structure of $S^n$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $Xsimeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Ysimeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Ysimeq Y/B$, and $Y/B$ is homeomorphic to $S^nvee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $simeq$ you are done.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $Xsimeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Ysimeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Ysimeq Y/B$, and $Y/B$ is homeomorphic to $S^nvee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $simeq$ you are done.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $Xsimeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Ysimeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Ysimeq Y/B$, and $Y/B$ is homeomorphic to $S^nvee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $simeq$ you are done.






          share|cite|improve this answer









          $endgroup$



          We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $Xsimeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Ysimeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Ysimeq Y/B$, and $Y/B$ is homeomorphic to $S^nvee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $simeq$ you are done.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 7:24









          Cheerful ParsnipCheerful Parsnip

          21k23397




          21k23397























              1












              $begingroup$

              For the inspiration, you should see Example 0.8 in Hatcher's book.



              Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 vee S^1$.



              You can generalize his argument using $CW$-complex structure of $S^n$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                For the inspiration, you should see Example 0.8 in Hatcher's book.



                Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 vee S^1$.



                You can generalize his argument using $CW$-complex structure of $S^n$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  For the inspiration, you should see Example 0.8 in Hatcher's book.



                  Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 vee S^1$.



                  You can generalize his argument using $CW$-complex structure of $S^n$.






                  share|cite|improve this answer









                  $endgroup$



                  For the inspiration, you should see Example 0.8 in Hatcher's book.



                  Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 vee S^1$.



                  You can generalize his argument using $CW$-complex structure of $S^n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 '18 at 8:53







                  user621469





































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