Retraction of the Möbius strip to its boundary












6












$begingroup$


Prove that there is no retraction (i.e. continuous function constant on the codomain) $r: M rightarrow S^1 = partial M$ where $M$ is the Möbius strip.



I've tried to find a contradiction using $r_*$ homomorphism between the fundamental groups, but they are both $mathbb{Z}$ and nothing seems to go wrong...






algebraic-topology mobius-band







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    6












    $begingroup$


    Prove that there is no retraction (i.e. continuous function constant on the codomain) $r: M rightarrow S^1 = partial M$ where $M$ is the Möbius strip.



    I've tried to find a contradiction using $r_*$ homomorphism between the fundamental groups, but they are both $mathbb{Z}$ and nothing seems to go wrong...






    algebraic-topology mobius-band







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      1



      $begingroup$


      Prove that there is no retraction (i.e. continuous function constant on the codomain) $r: M rightarrow S^1 = partial M$ where $M$ is the Möbius strip.



      I've tried to find a contradiction using $r_*$ homomorphism between the fundamental groups, but they are both $mathbb{Z}$ and nothing seems to go wrong...






      algebraic-topology mobius-band







      share|cite|improve this question











      $endgroup$




      Prove that there is no retraction (i.e. continuous function constant on the codomain) $r: M rightarrow S^1 = partial M$ where $M$ is the Möbius strip.



      I've tried to find a contradiction using $r_*$ homomorphism between the fundamental groups, but they are both $mathbb{Z}$ and nothing seems to go wrong...






      algebraic-topology mobius-band




      algebraic-topology mobius-band






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 8 '15 at 18:23









      Najib Idrissi

      41.6k473140




      41.6k473140










      asked Sep 25 '12 at 21:10







      user32847





























          3 Answers
          3






          active

          oldest

          votes


















          10












          $begingroup$

          If $alphainpi_1(partial M)$ is a generator, its image $i_*(alpha)inpi_1(M)$ under the inclusion $i:partial Mto M$ is the square of an element of $pi_1(M)$, so that if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$. This is not so.



          (For all this to work, one has to pick a basepoint $x_0inpartial M$ and use it to compute both $pi_1(M)$ and $pi_1(partial M)$)






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
            $endgroup$
            – user32847
            Sep 26 '12 at 7:31






          • 2




            $begingroup$
            Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
            $endgroup$
            – Mariano Suárez-Álvarez
            Sep 26 '12 at 17:36






          • 2




            $begingroup$
            Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
            $endgroup$
            – fierydemon
            Mar 15 '16 at 1:06










          • $begingroup$
            @AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
            $endgroup$
            – B.A
            Feb 22 '18 at 5:26










          • $begingroup$
            @MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
            $endgroup$
            – PerelMan
            Feb 22 at 17:35





















          2












          $begingroup$

          For each $alphainpartial M$, let $gamma_alpha$ be the closed loop in $M$ that starts at $alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $alphamapstogamma_alpha$ is a homotopy -- in particular every $gamma_alpha$ has the same homotopy class.



          On the other hand, if $x$ and $y$ are antipodes, then when we form $gamma_x+gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(gamma_x+gamma_y)$ in $partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $mathbb Z$, which is a contradiction.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (Annoyingly, one has to do this with based loops :-/ )
            $endgroup$
            – Mariano Suárez-Álvarez
            Sep 25 '12 at 21:39










          • $begingroup$
            @Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
            $endgroup$
            – Henning Makholm
            Sep 25 '12 at 21:43












          • $begingroup$
            @HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
            $endgroup$
            – PerelMan
            Feb 22 at 17:42










          • $begingroup$
            @PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
            $endgroup$
            – Henning Makholm
            Feb 22 at 18:42





















          1












          $begingroup$

          You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:Mrightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*alpha=i_*beta$). Thus we have an exact sequence
          $$
          0rightarrow H_1(B)xrightarrow{i_*}H_1(M)xrightarrow{q_*} H_1(M,B)rightarrow 0
          $$
          coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)cong H_1(M/B)$. But $M/B=mathbb Rmathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:



          enter image description here



          thus $H_1(X,B)cong mathbb Z/2mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.






          share|cite|improve this answer











          $endgroup$













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            3 Answers
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            3 Answers
            3






            active

            oldest

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            active

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            active

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            10












            $begingroup$

            If $alphainpi_1(partial M)$ is a generator, its image $i_*(alpha)inpi_1(M)$ under the inclusion $i:partial Mto M$ is the square of an element of $pi_1(M)$, so that if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$. This is not so.



            (For all this to work, one has to pick a basepoint $x_0inpartial M$ and use it to compute both $pi_1(M)$ and $pi_1(partial M)$)






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
              $endgroup$
              – user32847
              Sep 26 '12 at 7:31






            • 2




              $begingroup$
              Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
              $endgroup$
              – Mariano Suárez-Álvarez
              Sep 26 '12 at 17:36






            • 2




              $begingroup$
              Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
              $endgroup$
              – fierydemon
              Mar 15 '16 at 1:06










            • $begingroup$
              @AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
              $endgroup$
              – B.A
              Feb 22 '18 at 5:26










            • $begingroup$
              @MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
              $endgroup$
              – PerelMan
              Feb 22 at 17:35


















            10












            $begingroup$

            If $alphainpi_1(partial M)$ is a generator, its image $i_*(alpha)inpi_1(M)$ under the inclusion $i:partial Mto M$ is the square of an element of $pi_1(M)$, so that if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$. This is not so.



            (For all this to work, one has to pick a basepoint $x_0inpartial M$ and use it to compute both $pi_1(M)$ and $pi_1(partial M)$)






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
              $endgroup$
              – user32847
              Sep 26 '12 at 7:31






            • 2




              $begingroup$
              Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
              $endgroup$
              – Mariano Suárez-Álvarez
              Sep 26 '12 at 17:36






            • 2




              $begingroup$
              Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
              $endgroup$
              – fierydemon
              Mar 15 '16 at 1:06










            • $begingroup$
              @AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
              $endgroup$
              – B.A
              Feb 22 '18 at 5:26










            • $begingroup$
              @MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
              $endgroup$
              – PerelMan
              Feb 22 at 17:35
















            10












            10








            10





            $begingroup$

            If $alphainpi_1(partial M)$ is a generator, its image $i_*(alpha)inpi_1(M)$ under the inclusion $i:partial Mto M$ is the square of an element of $pi_1(M)$, so that if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$. This is not so.



            (For all this to work, one has to pick a basepoint $x_0inpartial M$ and use it to compute both $pi_1(M)$ and $pi_1(partial M)$)






            share|cite|improve this answer









            $endgroup$



            If $alphainpi_1(partial M)$ is a generator, its image $i_*(alpha)inpi_1(M)$ under the inclusion $i:partial Mto M$ is the square of an element of $pi_1(M)$, so that if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$. This is not so.



            (For all this to work, one has to pick a basepoint $x_0inpartial M$ and use it to compute both $pi_1(M)$ and $pi_1(partial M)$)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 25 '12 at 21:24









            Mariano Suárez-ÁlvarezMariano Suárez-Álvarez

            111k7157288




            111k7157288








            • 1




              $begingroup$
              why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
              $endgroup$
              – user32847
              Sep 26 '12 at 7:31






            • 2




              $begingroup$
              Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
              $endgroup$
              – Mariano Suárez-Álvarez
              Sep 26 '12 at 17:36






            • 2




              $begingroup$
              Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
              $endgroup$
              – fierydemon
              Mar 15 '16 at 1:06










            • $begingroup$
              @AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
              $endgroup$
              – B.A
              Feb 22 '18 at 5:26










            • $begingroup$
              @MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
              $endgroup$
              – PerelMan
              Feb 22 at 17:35
















            • 1




              $begingroup$
              why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
              $endgroup$
              – user32847
              Sep 26 '12 at 7:31






            • 2




              $begingroup$
              Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
              $endgroup$
              – Mariano Suárez-Álvarez
              Sep 26 '12 at 17:36






            • 2




              $begingroup$
              Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
              $endgroup$
              – fierydemon
              Mar 15 '16 at 1:06










            • $begingroup$
              @AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
              $endgroup$
              – B.A
              Feb 22 '18 at 5:26










            • $begingroup$
              @MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
              $endgroup$
              – PerelMan
              Feb 22 at 17:35










            1




            1




            $begingroup$
            why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
            $endgroup$
            – user32847
            Sep 26 '12 at 7:31




            $begingroup$
            why $i_* (alpha) $ is the square of an element in $pi_1 (M)$?
            $endgroup$
            – user32847
            Sep 26 '12 at 7:31




            2




            2




            $begingroup$
            Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
            $endgroup$
            – Mariano Suárez-Álvarez
            Sep 26 '12 at 17:36




            $begingroup$
            Essentially because it turns around the band twice, and a generator of the fundamental group of the band is a curve which turns around it only once. Notice that $i$ is a map one knows, so one can completely compute the morphism $i_*:pi_1(partial M,x_0)topi_1(M,x_0)$.
            $endgroup$
            – Mariano Suárez-Álvarez
            Sep 26 '12 at 17:36




            2




            2




            $begingroup$
            Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
            $endgroup$
            – fierydemon
            Mar 15 '16 at 1:06




            $begingroup$
            Why is the following true: "if $r:Mtopartial M$ is a retraction, $alpha=r_*i_*(alpha)$ is also the square of an element of $pi_1(partial M)$"?
            $endgroup$
            – fierydemon
            Mar 15 '16 at 1:06












            $begingroup$
            @AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
            $endgroup$
            – B.A
            Feb 22 '18 at 5:26




            $begingroup$
            @AyushKhaitan this is because $i_*(alpha) = beta^2$ and as we are dealing with homomorphism this means $r_*i_*(alpha)$ = r_*(beta)^2$
            $endgroup$
            – B.A
            Feb 22 '18 at 5:26












            $begingroup$
            @MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
            $endgroup$
            – PerelMan
            Feb 22 at 17:35






            $begingroup$
            @MarianoSuárez-Álvarez what is the point of computing $pi_1(M)$ and $pi_1(partial M)$? even if there is retraction, it is not a homotopy equivalence. Can we find a contradiction by these computations? Thanks!
            $endgroup$
            – PerelMan
            Feb 22 at 17:35













            2












            $begingroup$

            For each $alphainpartial M$, let $gamma_alpha$ be the closed loop in $M$ that starts at $alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $alphamapstogamma_alpha$ is a homotopy -- in particular every $gamma_alpha$ has the same homotopy class.



            On the other hand, if $x$ and $y$ are antipodes, then when we form $gamma_x+gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(gamma_x+gamma_y)$ in $partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $mathbb Z$, which is a contradiction.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              (Annoyingly, one has to do this with based loops :-/ )
              $endgroup$
              – Mariano Suárez-Álvarez
              Sep 25 '12 at 21:39










            • $begingroup$
              @Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
              $endgroup$
              – Henning Makholm
              Sep 25 '12 at 21:43












            • $begingroup$
              @HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
              $endgroup$
              – PerelMan
              Feb 22 at 17:42










            • $begingroup$
              @PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
              $endgroup$
              – Henning Makholm
              Feb 22 at 18:42


















            2












            $begingroup$

            For each $alphainpartial M$, let $gamma_alpha$ be the closed loop in $M$ that starts at $alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $alphamapstogamma_alpha$ is a homotopy -- in particular every $gamma_alpha$ has the same homotopy class.



            On the other hand, if $x$ and $y$ are antipodes, then when we form $gamma_x+gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(gamma_x+gamma_y)$ in $partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $mathbb Z$, which is a contradiction.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              (Annoyingly, one has to do this with based loops :-/ )
              $endgroup$
              – Mariano Suárez-Álvarez
              Sep 25 '12 at 21:39










            • $begingroup$
              @Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
              $endgroup$
              – Henning Makholm
              Sep 25 '12 at 21:43












            • $begingroup$
              @HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
              $endgroup$
              – PerelMan
              Feb 22 at 17:42










            • $begingroup$
              @PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
              $endgroup$
              – Henning Makholm
              Feb 22 at 18:42
















            2












            2








            2





            $begingroup$

            For each $alphainpartial M$, let $gamma_alpha$ be the closed loop in $M$ that starts at $alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $alphamapstogamma_alpha$ is a homotopy -- in particular every $gamma_alpha$ has the same homotopy class.



            On the other hand, if $x$ and $y$ are antipodes, then when we form $gamma_x+gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(gamma_x+gamma_y)$ in $partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $mathbb Z$, which is a contradiction.






            share|cite|improve this answer









            $endgroup$



            For each $alphainpartial M$, let $gamma_alpha$ be the closed loop in $M$ that starts at $alpha$, goes directly across the strip to its antipode and then halfway around the boundary to its starting point in positive direction. Then $alphamapstogamma_alpha$ is a homotopy -- in particular every $gamma_alpha$ has the same homotopy class.



            On the other hand, if $x$ and $y$ are antipodes, then when we form $gamma_x+gamma_y$, the "directly across" sections cancel out, and the concatenated curve is homotopic to a single turn around the entire boundary. So the homotopy class of $r(gamma_x+gamma_y)$ in $partial M$ is $1$. On the other hand, $r$ ought to induce a homomorphism between the homotopy groups, but $1$ is not twice anything in $mathbb Z$, which is a contradiction.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 25 '12 at 21:37









            Henning MakholmHenning Makholm

            241k17308546




            241k17308546












            • $begingroup$
              (Annoyingly, one has to do this with based loops :-/ )
              $endgroup$
              – Mariano Suárez-Álvarez
              Sep 25 '12 at 21:39










            • $begingroup$
              @Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
              $endgroup$
              – Henning Makholm
              Sep 25 '12 at 21:43












            • $begingroup$
              @HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
              $endgroup$
              – PerelMan
              Feb 22 at 17:42










            • $begingroup$
              @PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
              $endgroup$
              – Henning Makholm
              Feb 22 at 18:42




















            • $begingroup$
              (Annoyingly, one has to do this with based loops :-/ )
              $endgroup$
              – Mariano Suárez-Álvarez
              Sep 25 '12 at 21:39










            • $begingroup$
              @Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
              $endgroup$
              – Henning Makholm
              Sep 25 '12 at 21:43












            • $begingroup$
              @HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
              $endgroup$
              – PerelMan
              Feb 22 at 17:42










            • $begingroup$
              @PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
              $endgroup$
              – Henning Makholm
              Feb 22 at 18:42


















            $begingroup$
            (Annoyingly, one has to do this with based loops :-/ )
            $endgroup$
            – Mariano Suárez-Álvarez
            Sep 25 '12 at 21:39




            $begingroup$
            (Annoyingly, one has to do this with based loops :-/ )
            $endgroup$
            – Mariano Suárez-Álvarez
            Sep 25 '12 at 21:39












            $begingroup$
            @Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
            $endgroup$
            – Henning Makholm
            Sep 25 '12 at 21:43






            $begingroup$
            @Mariano: Reparameterizing all of the loops to be based at $x$ and then throwing away the $alpha$s where $gamma_alpha$ does not already pass through $x$ ought to take care of that.
            $endgroup$
            – Henning Makholm
            Sep 25 '12 at 21:43














            $begingroup$
            @HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
            $endgroup$
            – PerelMan
            Feb 22 at 17:42




            $begingroup$
            @HenningMakholm is $pi_1(partial M)=1$ or I misunderstood something?
            $endgroup$
            – PerelMan
            Feb 22 at 17:42












            $begingroup$
            @PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
            $endgroup$
            – Henning Makholm
            Feb 22 at 18:42






            $begingroup$
            @PerelMan: $partial M$ is a circle, so $pi_1(partial M)congmathbb Z$. When I write $1$ I mean $1inmathbb Z$.
            $endgroup$
            – Henning Makholm
            Feb 22 at 18:42













            1












            $begingroup$

            You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:Mrightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*alpha=i_*beta$). Thus we have an exact sequence
            $$
            0rightarrow H_1(B)xrightarrow{i_*}H_1(M)xrightarrow{q_*} H_1(M,B)rightarrow 0
            $$
            coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)cong H_1(M/B)$. But $M/B=mathbb Rmathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:



            enter image description here



            thus $H_1(X,B)cong mathbb Z/2mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:Mrightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*alpha=i_*beta$). Thus we have an exact sequence
              $$
              0rightarrow H_1(B)xrightarrow{i_*}H_1(M)xrightarrow{q_*} H_1(M,B)rightarrow 0
              $$
              coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)cong H_1(M/B)$. But $M/B=mathbb Rmathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:



              enter image description here



              thus $H_1(X,B)cong mathbb Z/2mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:Mrightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*alpha=i_*beta$). Thus we have an exact sequence
                $$
                0rightarrow H_1(B)xrightarrow{i_*}H_1(M)xrightarrow{q_*} H_1(M,B)rightarrow 0
                $$
                coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)cong H_1(M/B)$. But $M/B=mathbb Rmathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:



                enter image description here



                thus $H_1(X,B)cong mathbb Z/2mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.






                share|cite|improve this answer











                $endgroup$



                You can also prove this using homology, but it's somewhat more effort. The basic idea is that, if $B$ is the boundary circle and $r:Mrightarrow M$ is a retraction onto $B$ then the inclusion map also induces an injection $i_*:H_1(B)rightarrow H_1(X)$ (to see this, apply $r_*$ to $i_*alpha=i_*beta$). Thus we have an exact sequence
                $$
                0rightarrow H_1(B)xrightarrow{i_*}H_1(M)xrightarrow{q_*} H_1(M,B)rightarrow 0
                $$
                coming from the reduced long exact sequence for the pair $(M,B)$. We know that $H_1(B)$ and $H_1(M)$ are both $mathbb Z$ (because $B=S^1$ and $M$ deformation retracts onto its central circle) and, since $(M,B)$ is a good pair, $H_1(M,B)cong H_1(M/B)$. But $M/B=mathbb Rmathbb P^2$, as can be seen by their cell decompositions, where the pink indicates the boundary circle $B$:



                enter image description here



                thus $H_1(X,B)cong mathbb Z/2mathbb Z$. The fact that $r$ is a retraction means that the above sequence splits, as $r_*:H_2(M)rightarrow H_2(B)$ composed with $i_*$ is the identity. This is a contradiction.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 10 at 15:54

























                answered Jan 10 at 4:06









                ArbutusArbutus

                701715




                701715






























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