What is the space by identifying the antipodal points of a cylinder?












1












$begingroup$


Define the equivalence relation for a cylinder $S^1times [-1,1]subset mathbb{R}^3$ as $(x,y,z) sim (-x,-y,-z)$. We know that identifying antipodal points of $S^2$ or of the boundary of the disk $D^1$ gives projective plane $mathbb{RP}^2$, I wonder if there a simple surface for identifying the antipodal points of a cylinder? Is it $mathbb{RP}^2$?






general-topology







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$endgroup$








  • 1




    $begingroup$
    $S^1 times [-1,1]$ is two-dimensional and $mathbb{RP}^3$ is three-dimensional.
    $endgroup$
    – Paul Frost
    Jan 20 at 14:29








  • 1




    $begingroup$
    Could you suggest what this space is? @PaulFrost
    $endgroup$
    – Nicky
    Jan 20 at 15:00












  • $begingroup$
    I would say your surface is related to the Klein bottle but I'm not sure at all.
    $endgroup$
    – Dog_69
    Jan 20 at 15:09
















1












$begingroup$


Define the equivalence relation for a cylinder $S^1times [-1,1]subset mathbb{R}^3$ as $(x,y,z) sim (-x,-y,-z)$. We know that identifying antipodal points of $S^2$ or of the boundary of the disk $D^1$ gives projective plane $mathbb{RP}^2$, I wonder if there a simple surface for identifying the antipodal points of a cylinder? Is it $mathbb{RP}^2$?






general-topology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $S^1 times [-1,1]$ is two-dimensional and $mathbb{RP}^3$ is three-dimensional.
    $endgroup$
    – Paul Frost
    Jan 20 at 14:29








  • 1




    $begingroup$
    Could you suggest what this space is? @PaulFrost
    $endgroup$
    – Nicky
    Jan 20 at 15:00












  • $begingroup$
    I would say your surface is related to the Klein bottle but I'm not sure at all.
    $endgroup$
    – Dog_69
    Jan 20 at 15:09














1












1








1


2



$begingroup$


Define the equivalence relation for a cylinder $S^1times [-1,1]subset mathbb{R}^3$ as $(x,y,z) sim (-x,-y,-z)$. We know that identifying antipodal points of $S^2$ or of the boundary of the disk $D^1$ gives projective plane $mathbb{RP}^2$, I wonder if there a simple surface for identifying the antipodal points of a cylinder? Is it $mathbb{RP}^2$?






general-topology







share|cite|improve this question











$endgroup$




Define the equivalence relation for a cylinder $S^1times [-1,1]subset mathbb{R}^3$ as $(x,y,z) sim (-x,-y,-z)$. We know that identifying antipodal points of $S^2$ or of the boundary of the disk $D^1$ gives projective plane $mathbb{RP}^2$, I wonder if there a simple surface for identifying the antipodal points of a cylinder? Is it $mathbb{RP}^2$?






general-topology




general-topology






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 15:14







Nicky

















asked Jan 20 at 14:16









NickyNicky

465




465








  • 1




    $begingroup$
    $S^1 times [-1,1]$ is two-dimensional and $mathbb{RP}^3$ is three-dimensional.
    $endgroup$
    – Paul Frost
    Jan 20 at 14:29








  • 1




    $begingroup$
    Could you suggest what this space is? @PaulFrost
    $endgroup$
    – Nicky
    Jan 20 at 15:00












  • $begingroup$
    I would say your surface is related to the Klein bottle but I'm not sure at all.
    $endgroup$
    – Dog_69
    Jan 20 at 15:09














  • 1




    $begingroup$
    $S^1 times [-1,1]$ is two-dimensional and $mathbb{RP}^3$ is three-dimensional.
    $endgroup$
    – Paul Frost
    Jan 20 at 14:29








  • 1




    $begingroup$
    Could you suggest what this space is? @PaulFrost
    $endgroup$
    – Nicky
    Jan 20 at 15:00












  • $begingroup$
    I would say your surface is related to the Klein bottle but I'm not sure at all.
    $endgroup$
    – Dog_69
    Jan 20 at 15:09








1




1




$begingroup$
$S^1 times [-1,1]$ is two-dimensional and $mathbb{RP}^3$ is three-dimensional.
$endgroup$
– Paul Frost
Jan 20 at 14:29






$begingroup$
$S^1 times [-1,1]$ is two-dimensional and $mathbb{RP}^3$ is three-dimensional.
$endgroup$
– Paul Frost
Jan 20 at 14:29






1




1




$begingroup$
Could you suggest what this space is? @PaulFrost
$endgroup$
– Nicky
Jan 20 at 15:00






$begingroup$
Could you suggest what this space is? @PaulFrost
$endgroup$
– Nicky
Jan 20 at 15:00














$begingroup$
I would say your surface is related to the Klein bottle but I'm not sure at all.
$endgroup$
– Dog_69
Jan 20 at 15:09




$begingroup$
I would say your surface is related to the Klein bottle but I'm not sure at all.
$endgroup$
– Dog_69
Jan 20 at 15:09










2 Answers
2






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0












$begingroup$

Let $S^1 = { z in mathbb{C} = mathbb{R}^2 mid lvert z rvert = 1 }$. Define $q : S^1 to S^1, q(z) = z^2$. Let $C(q)$ denote the mapping cylinder of $q$ which is the quotient space obtained from the disjoint union $S^1 times [0,1] + S^1$ with respect to the equivalence relation generated by $(z,0) sim q(z)$. Let $r : S^1 times [0,1] + S^1 to C(q)$ denote the quotient map. Note that $rho = r mid_{S^1 times [0,1]}$ is surjective because $q$ is surjective.



Define $phi : S^1 times [-1,0] to S^1 times [0,1], phi(z,t) = -(z,t)$. Then $rho(z,0) = rho(phi(z,0))$ because $(z,0) sim z^2 = (-z)^2 sim (-z,0) = phi(z,0)$. Next define
$$f : S^1 times [-1,1] to C(q), f(z,t) =
begin{cases}
rho(z,t) & t ge 0 \
rho(phi(z,t)) & t le 0
end{cases}
$$
This is a continuous surjection (note that both parts of the defnition agree on $S^1 times [0,1] cap S^1 times [-1,0] = S^1 times { 0 }$).



We have $f(z,t) = f(w,s)$ if and only if $(w,s) = pm(z,t)$. The "if"-part is trivial. So let $f(z,t) = f(w,s)$. If $s = 0$, we must have $t = 0$ and $z^2 = w^2$, i.e. $w = pm z$. If $s ne 0$, we must have $t = epsilon s$ with $epsilon = pm 1$. But then clearly $w = epsilon z$.



Hence $S^1 times [-1,1]/sim$ is homeomorphic to $C(q)$. The latter is homotopy equivalent to $S^1$.



Note that the space $C(q)$ is homeomorphic to $mathbb{RP}^2 setminus O$ where $O$ is an open disk in $mathbb{RP}^2$.






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    0












    $begingroup$

    Here is another answer. Let $S = S^1 times [-1,1] cup D^2 times { -1, 1 }$. There is a homeomorphism $h : S to S^2$ such that $h circ a = a circ h$, where $a$ denotes the antipodal map both on $S$ and $S^2$. Hence $S/ sim$, where $xi sim a(xi)$, is homeomorphic to $mathbb{RP}^2$. Let $p : S to R = S/sim$ denote the quotient map and $U = mathring{D} times { -1, 1 }$. Then $O = p(U)$ is an open disk in $R$ and $p$ restricts to a quotient map $p ' : S^1 times[-1,1] = S setminus U to R setminus O$.






    share|cite|improve this answer









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      2 Answers
      2






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      2 Answers
      2






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      active

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      active

      oldest

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      0












      $begingroup$

      Let $S^1 = { z in mathbb{C} = mathbb{R}^2 mid lvert z rvert = 1 }$. Define $q : S^1 to S^1, q(z) = z^2$. Let $C(q)$ denote the mapping cylinder of $q$ which is the quotient space obtained from the disjoint union $S^1 times [0,1] + S^1$ with respect to the equivalence relation generated by $(z,0) sim q(z)$. Let $r : S^1 times [0,1] + S^1 to C(q)$ denote the quotient map. Note that $rho = r mid_{S^1 times [0,1]}$ is surjective because $q$ is surjective.



      Define $phi : S^1 times [-1,0] to S^1 times [0,1], phi(z,t) = -(z,t)$. Then $rho(z,0) = rho(phi(z,0))$ because $(z,0) sim z^2 = (-z)^2 sim (-z,0) = phi(z,0)$. Next define
      $$f : S^1 times [-1,1] to C(q), f(z,t) =
      begin{cases}
      rho(z,t) & t ge 0 \
      rho(phi(z,t)) & t le 0
      end{cases}
      $$
      This is a continuous surjection (note that both parts of the defnition agree on $S^1 times [0,1] cap S^1 times [-1,0] = S^1 times { 0 }$).



      We have $f(z,t) = f(w,s)$ if and only if $(w,s) = pm(z,t)$. The "if"-part is trivial. So let $f(z,t) = f(w,s)$. If $s = 0$, we must have $t = 0$ and $z^2 = w^2$, i.e. $w = pm z$. If $s ne 0$, we must have $t = epsilon s$ with $epsilon = pm 1$. But then clearly $w = epsilon z$.



      Hence $S^1 times [-1,1]/sim$ is homeomorphic to $C(q)$. The latter is homotopy equivalent to $S^1$.



      Note that the space $C(q)$ is homeomorphic to $mathbb{RP}^2 setminus O$ where $O$ is an open disk in $mathbb{RP}^2$.






      share|cite|improve this answer











      $endgroup$


















        0












        $begingroup$

        Let $S^1 = { z in mathbb{C} = mathbb{R}^2 mid lvert z rvert = 1 }$. Define $q : S^1 to S^1, q(z) = z^2$. Let $C(q)$ denote the mapping cylinder of $q$ which is the quotient space obtained from the disjoint union $S^1 times [0,1] + S^1$ with respect to the equivalence relation generated by $(z,0) sim q(z)$. Let $r : S^1 times [0,1] + S^1 to C(q)$ denote the quotient map. Note that $rho = r mid_{S^1 times [0,1]}$ is surjective because $q$ is surjective.



        Define $phi : S^1 times [-1,0] to S^1 times [0,1], phi(z,t) = -(z,t)$. Then $rho(z,0) = rho(phi(z,0))$ because $(z,0) sim z^2 = (-z)^2 sim (-z,0) = phi(z,0)$. Next define
        $$f : S^1 times [-1,1] to C(q), f(z,t) =
        begin{cases}
        rho(z,t) & t ge 0 \
        rho(phi(z,t)) & t le 0
        end{cases}
        $$
        This is a continuous surjection (note that both parts of the defnition agree on $S^1 times [0,1] cap S^1 times [-1,0] = S^1 times { 0 }$).



        We have $f(z,t) = f(w,s)$ if and only if $(w,s) = pm(z,t)$. The "if"-part is trivial. So let $f(z,t) = f(w,s)$. If $s = 0$, we must have $t = 0$ and $z^2 = w^2$, i.e. $w = pm z$. If $s ne 0$, we must have $t = epsilon s$ with $epsilon = pm 1$. But then clearly $w = epsilon z$.



        Hence $S^1 times [-1,1]/sim$ is homeomorphic to $C(q)$. The latter is homotopy equivalent to $S^1$.



        Note that the space $C(q)$ is homeomorphic to $mathbb{RP}^2 setminus O$ where $O$ is an open disk in $mathbb{RP}^2$.






        share|cite|improve this answer











        $endgroup$
















          0












          0








          0





          $begingroup$

          Let $S^1 = { z in mathbb{C} = mathbb{R}^2 mid lvert z rvert = 1 }$. Define $q : S^1 to S^1, q(z) = z^2$. Let $C(q)$ denote the mapping cylinder of $q$ which is the quotient space obtained from the disjoint union $S^1 times [0,1] + S^1$ with respect to the equivalence relation generated by $(z,0) sim q(z)$. Let $r : S^1 times [0,1] + S^1 to C(q)$ denote the quotient map. Note that $rho = r mid_{S^1 times [0,1]}$ is surjective because $q$ is surjective.



          Define $phi : S^1 times [-1,0] to S^1 times [0,1], phi(z,t) = -(z,t)$. Then $rho(z,0) = rho(phi(z,0))$ because $(z,0) sim z^2 = (-z)^2 sim (-z,0) = phi(z,0)$. Next define
          $$f : S^1 times [-1,1] to C(q), f(z,t) =
          begin{cases}
          rho(z,t) & t ge 0 \
          rho(phi(z,t)) & t le 0
          end{cases}
          $$
          This is a continuous surjection (note that both parts of the defnition agree on $S^1 times [0,1] cap S^1 times [-1,0] = S^1 times { 0 }$).



          We have $f(z,t) = f(w,s)$ if and only if $(w,s) = pm(z,t)$. The "if"-part is trivial. So let $f(z,t) = f(w,s)$. If $s = 0$, we must have $t = 0$ and $z^2 = w^2$, i.e. $w = pm z$. If $s ne 0$, we must have $t = epsilon s$ with $epsilon = pm 1$. But then clearly $w = epsilon z$.



          Hence $S^1 times [-1,1]/sim$ is homeomorphic to $C(q)$. The latter is homotopy equivalent to $S^1$.



          Note that the space $C(q)$ is homeomorphic to $mathbb{RP}^2 setminus O$ where $O$ is an open disk in $mathbb{RP}^2$.






          share|cite|improve this answer











          $endgroup$



          Let $S^1 = { z in mathbb{C} = mathbb{R}^2 mid lvert z rvert = 1 }$. Define $q : S^1 to S^1, q(z) = z^2$. Let $C(q)$ denote the mapping cylinder of $q$ which is the quotient space obtained from the disjoint union $S^1 times [0,1] + S^1$ with respect to the equivalence relation generated by $(z,0) sim q(z)$. Let $r : S^1 times [0,1] + S^1 to C(q)$ denote the quotient map. Note that $rho = r mid_{S^1 times [0,1]}$ is surjective because $q$ is surjective.



          Define $phi : S^1 times [-1,0] to S^1 times [0,1], phi(z,t) = -(z,t)$. Then $rho(z,0) = rho(phi(z,0))$ because $(z,0) sim z^2 = (-z)^2 sim (-z,0) = phi(z,0)$. Next define
          $$f : S^1 times [-1,1] to C(q), f(z,t) =
          begin{cases}
          rho(z,t) & t ge 0 \
          rho(phi(z,t)) & t le 0
          end{cases}
          $$
          This is a continuous surjection (note that both parts of the defnition agree on $S^1 times [0,1] cap S^1 times [-1,0] = S^1 times { 0 }$).



          We have $f(z,t) = f(w,s)$ if and only if $(w,s) = pm(z,t)$. The "if"-part is trivial. So let $f(z,t) = f(w,s)$. If $s = 0$, we must have $t = 0$ and $z^2 = w^2$, i.e. $w = pm z$. If $s ne 0$, we must have $t = epsilon s$ with $epsilon = pm 1$. But then clearly $w = epsilon z$.



          Hence $S^1 times [-1,1]/sim$ is homeomorphic to $C(q)$. The latter is homotopy equivalent to $S^1$.



          Note that the space $C(q)$ is homeomorphic to $mathbb{RP}^2 setminus O$ where $O$ is an open disk in $mathbb{RP}^2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 20 at 23:21

























          answered Jan 20 at 15:48









          Paul FrostPaul Frost

          11k3934




          11k3934























              0












              $begingroup$

              Here is another answer. Let $S = S^1 times [-1,1] cup D^2 times { -1, 1 }$. There is a homeomorphism $h : S to S^2$ such that $h circ a = a circ h$, where $a$ denotes the antipodal map both on $S$ and $S^2$. Hence $S/ sim$, where $xi sim a(xi)$, is homeomorphic to $mathbb{RP}^2$. Let $p : S to R = S/sim$ denote the quotient map and $U = mathring{D} times { -1, 1 }$. Then $O = p(U)$ is an open disk in $R$ and $p$ restricts to a quotient map $p ' : S^1 times[-1,1] = S setminus U to R setminus O$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Here is another answer. Let $S = S^1 times [-1,1] cup D^2 times { -1, 1 }$. There is a homeomorphism $h : S to S^2$ such that $h circ a = a circ h$, where $a$ denotes the antipodal map both on $S$ and $S^2$. Hence $S/ sim$, where $xi sim a(xi)$, is homeomorphic to $mathbb{RP}^2$. Let $p : S to R = S/sim$ denote the quotient map and $U = mathring{D} times { -1, 1 }$. Then $O = p(U)$ is an open disk in $R$ and $p$ restricts to a quotient map $p ' : S^1 times[-1,1] = S setminus U to R setminus O$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Here is another answer. Let $S = S^1 times [-1,1] cup D^2 times { -1, 1 }$. There is a homeomorphism $h : S to S^2$ such that $h circ a = a circ h$, where $a$ denotes the antipodal map both on $S$ and $S^2$. Hence $S/ sim$, where $xi sim a(xi)$, is homeomorphic to $mathbb{RP}^2$. Let $p : S to R = S/sim$ denote the quotient map and $U = mathring{D} times { -1, 1 }$. Then $O = p(U)$ is an open disk in $R$ and $p$ restricts to a quotient map $p ' : S^1 times[-1,1] = S setminus U to R setminus O$.






                  share|cite|improve this answer









                  $endgroup$



                  Here is another answer. Let $S = S^1 times [-1,1] cup D^2 times { -1, 1 }$. There is a homeomorphism $h : S to S^2$ such that $h circ a = a circ h$, where $a$ denotes the antipodal map both on $S$ and $S^2$. Hence $S/ sim$, where $xi sim a(xi)$, is homeomorphic to $mathbb{RP}^2$. Let $p : S to R = S/sim$ denote the quotient map and $U = mathring{D} times { -1, 1 }$. Then $O = p(U)$ is an open disk in $R$ and $p$ restricts to a quotient map $p ' : S^1 times[-1,1] = S setminus U to R setminus O$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 22 at 16:26









                  Paul FrostPaul Frost

                  11k3934




                  11k3934






























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