Same homotopy type as circle












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$begingroup$


Let $S={(z_1,z_2,ldots, z_n)in mathbb{C}^n: |z_1|^2+|z_2|^2+ldots+|z_n|^2=1}$ let $K={(z_1,z_2,ldots, z_n)in S: |z_1|^2+|z_2|^2+ldots+|z_{n-1}|^2=1}$. Prove that $Ssetminus K=M$ has same homotopy type as circle?



My attempt: We define the deformation retract of $M$ to ${(z_1,z_2,ldots, z_n)in S:|z_n|^2=1}subset M$ by



$$f_t=frac{(1-t)(z_1,z_2,ldots,z_n)+ t(0,0,ldots, 0,frac{z_{n}}{|z_n|} )}{|(1-t)(z_1,z_2,ldots,z_n)+ t(0,0,ldots, 0,frac{z_{n}}{|z_n|} )|}$$



Will this homotopy work?



Are there any other ways to prove the result?






proof-verification algebraic-topology differential-topology singularity-theory







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    0












    $begingroup$


    Let $S={(z_1,z_2,ldots, z_n)in mathbb{C}^n: |z_1|^2+|z_2|^2+ldots+|z_n|^2=1}$ let $K={(z_1,z_2,ldots, z_n)in S: |z_1|^2+|z_2|^2+ldots+|z_{n-1}|^2=1}$. Prove that $Ssetminus K=M$ has same homotopy type as circle?



    My attempt: We define the deformation retract of $M$ to ${(z_1,z_2,ldots, z_n)in S:|z_n|^2=1}subset M$ by



    $$f_t=frac{(1-t)(z_1,z_2,ldots,z_n)+ t(0,0,ldots, 0,frac{z_{n}}{|z_n|} )}{|(1-t)(z_1,z_2,ldots,z_n)+ t(0,0,ldots, 0,frac{z_{n}}{|z_n|} )|}$$



    Will this homotopy work?



    Are there any other ways to prove the result?






    proof-verification algebraic-topology differential-topology singularity-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $S={(z_1,z_2,ldots, z_n)in mathbb{C}^n: |z_1|^2+|z_2|^2+ldots+|z_n|^2=1}$ let $K={(z_1,z_2,ldots, z_n)in S: |z_1|^2+|z_2|^2+ldots+|z_{n-1}|^2=1}$. Prove that $Ssetminus K=M$ has same homotopy type as circle?



      My attempt: We define the deformation retract of $M$ to ${(z_1,z_2,ldots, z_n)in S:|z_n|^2=1}subset M$ by



      $$f_t=frac{(1-t)(z_1,z_2,ldots,z_n)+ t(0,0,ldots, 0,frac{z_{n}}{|z_n|} )}{|(1-t)(z_1,z_2,ldots,z_n)+ t(0,0,ldots, 0,frac{z_{n}}{|z_n|} )|}$$



      Will this homotopy work?



      Are there any other ways to prove the result?






      proof-verification algebraic-topology differential-topology singularity-theory







      share|cite|improve this question









      $endgroup$




      Let $S={(z_1,z_2,ldots, z_n)in mathbb{C}^n: |z_1|^2+|z_2|^2+ldots+|z_n|^2=1}$ let $K={(z_1,z_2,ldots, z_n)in S: |z_1|^2+|z_2|^2+ldots+|z_{n-1}|^2=1}$. Prove that $Ssetminus K=M$ has same homotopy type as circle?



      My attempt: We define the deformation retract of $M$ to ${(z_1,z_2,ldots, z_n)in S:|z_n|^2=1}subset M$ by



      $$f_t=frac{(1-t)(z_1,z_2,ldots,z_n)+ t(0,0,ldots, 0,frac{z_{n}}{|z_n|} )}{|(1-t)(z_1,z_2,ldots,z_n)+ t(0,0,ldots, 0,frac{z_{n}}{|z_n|} )|}$$



      Will this homotopy work?



      Are there any other ways to prove the result?






      proof-verification algebraic-topology differential-topology singularity-theory




      proof-verification algebraic-topology differential-topology singularity-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 30 at 11:15









      user345777user345777

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      432312






















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          Yes it works. What details do you still need to check? Maybe:




          • This straight line homotopy in the numerator is never zero when $z_n neq 0$, so it does make sense to normalize it. This follows from the next point.


          • The homotopy stays in $Ssetminus K$, i.e. the last coordinate of the straight line homotopy is never zero if $z_n neq 0$ . To check this note $(1-t)z_n + tz_n/|z_n| = 0$ is equivalent to $1 = t(1-1/|z_n|)$ but $t(1-1/|z_n|) leq (1-1/|z_n|) leq 1$







          share|cite|improve this answer









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            $begingroup$

            Yes it works. What details do you still need to check? Maybe:




            • This straight line homotopy in the numerator is never zero when $z_n neq 0$, so it does make sense to normalize it. This follows from the next point.


            • The homotopy stays in $Ssetminus K$, i.e. the last coordinate of the straight line homotopy is never zero if $z_n neq 0$ . To check this note $(1-t)z_n + tz_n/|z_n| = 0$ is equivalent to $1 = t(1-1/|z_n|)$ but $t(1-1/|z_n|) leq (1-1/|z_n|) leq 1$







            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Yes it works. What details do you still need to check? Maybe:




              • This straight line homotopy in the numerator is never zero when $z_n neq 0$, so it does make sense to normalize it. This follows from the next point.


              • The homotopy stays in $Ssetminus K$, i.e. the last coordinate of the straight line homotopy is never zero if $z_n neq 0$ . To check this note $(1-t)z_n + tz_n/|z_n| = 0$ is equivalent to $1 = t(1-1/|z_n|)$ but $t(1-1/|z_n|) leq (1-1/|z_n|) leq 1$







              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Yes it works. What details do you still need to check? Maybe:




                • This straight line homotopy in the numerator is never zero when $z_n neq 0$, so it does make sense to normalize it. This follows from the next point.


                • The homotopy stays in $Ssetminus K$, i.e. the last coordinate of the straight line homotopy is never zero if $z_n neq 0$ . To check this note $(1-t)z_n + tz_n/|z_n| = 0$ is equivalent to $1 = t(1-1/|z_n|)$ but $t(1-1/|z_n|) leq (1-1/|z_n|) leq 1$







                share|cite|improve this answer









                $endgroup$



                Yes it works. What details do you still need to check? Maybe:




                • This straight line homotopy in the numerator is never zero when $z_n neq 0$, so it does make sense to normalize it. This follows from the next point.


                • The homotopy stays in $Ssetminus K$, i.e. the last coordinate of the straight line homotopy is never zero if $z_n neq 0$ . To check this note $(1-t)z_n + tz_n/|z_n| = 0$ is equivalent to $1 = t(1-1/|z_n|)$ but $t(1-1/|z_n|) leq (1-1/|z_n|) leq 1$








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 at 14:08









                BenBen

                4,318617




                4,318617






























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