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Fundamental group of $mathbb{R}P^2$ in 2 models


Fundamental group of the torusGood exercises to do/examples to illustrate Seifert - Van Kampen TheoremFundamental group of this spaceFundamental group of $S^{1}$ unioned with its two diametersFundamental Group of a Quotient of an AnnulusFundamental Group of Surface Formed by Gluing Three Mobius Strips along their BoundariesFundamental group of the sphere with n-points identifiedFundamental group of projective plane with g handles by van KampenFundamental group of complement spaceFundamental group of torus knot without thickening













1












$begingroup$


I know that $pi_1(mathbb{R}P^2)congmathbb{Z}_2$, but in the square model, I get that $pi_1(mathbb{R}P^2)=langle a,bcolon ababrangle$. These groups must be isomorphic, but I can't find the isomorphism. What is the trick?



$mathbb{R}P^2$ is the following model



enter image description here



We want to use Seifert van Kampen, so let $U$ be the complement of a point and $V$ an open ball around that point. Then $U$ deformation retracts onto the boundary which is homotopy equivalent to $S^1vee S^1$. $V$ is simply connected and $Ucap V$ deformation retracts on $S^1$. Then,



$pi_1(mathbb{R}P^2)congmathbb{Z}*mathbb{Z}*_mathbb{Z} 1$. Let $i:Ucap Vto U$ be the inclusion, Then $pi_1(mathbb{R}P^2)conglangle a,bcolon i_*(1)rangle=langle a,bcolon ababrangle$










share|cite|improve this question











$endgroup$












  • $begingroup$
    That's not the correct presentation; your second group maps onto $(mathbb{Z/2})^2$
    $endgroup$
    – Max
    Mar 11 at 11:08










  • $begingroup$
    Perhaps write down your "square model" and find the mistake.
    $endgroup$
    – Tyrone
    Mar 11 at 11:15










  • $begingroup$
    I have edited my question
    $endgroup$
    – James
    Mar 11 at 11:23










  • $begingroup$
    Is the mistake that $U$ doesn't deformation retract on the bouquet of circles?
    $endgroup$
    – James
    Mar 11 at 11:32
















1












$begingroup$


I know that $pi_1(mathbb{R}P^2)congmathbb{Z}_2$, but in the square model, I get that $pi_1(mathbb{R}P^2)=langle a,bcolon ababrangle$. These groups must be isomorphic, but I can't find the isomorphism. What is the trick?



$mathbb{R}P^2$ is the following model



enter image description here



We want to use Seifert van Kampen, so let $U$ be the complement of a point and $V$ an open ball around that point. Then $U$ deformation retracts onto the boundary which is homotopy equivalent to $S^1vee S^1$. $V$ is simply connected and $Ucap V$ deformation retracts on $S^1$. Then,



$pi_1(mathbb{R}P^2)congmathbb{Z}*mathbb{Z}*_mathbb{Z} 1$. Let $i:Ucap Vto U$ be the inclusion, Then $pi_1(mathbb{R}P^2)conglangle a,bcolon i_*(1)rangle=langle a,bcolon ababrangle$










share|cite|improve this question











$endgroup$












  • $begingroup$
    That's not the correct presentation; your second group maps onto $(mathbb{Z/2})^2$
    $endgroup$
    – Max
    Mar 11 at 11:08










  • $begingroup$
    Perhaps write down your "square model" and find the mistake.
    $endgroup$
    – Tyrone
    Mar 11 at 11:15










  • $begingroup$
    I have edited my question
    $endgroup$
    – James
    Mar 11 at 11:23










  • $begingroup$
    Is the mistake that $U$ doesn't deformation retract on the bouquet of circles?
    $endgroup$
    – James
    Mar 11 at 11:32














1












1








1





$begingroup$


I know that $pi_1(mathbb{R}P^2)congmathbb{Z}_2$, but in the square model, I get that $pi_1(mathbb{R}P^2)=langle a,bcolon ababrangle$. These groups must be isomorphic, but I can't find the isomorphism. What is the trick?



$mathbb{R}P^2$ is the following model



enter image description here



We want to use Seifert van Kampen, so let $U$ be the complement of a point and $V$ an open ball around that point. Then $U$ deformation retracts onto the boundary which is homotopy equivalent to $S^1vee S^1$. $V$ is simply connected and $Ucap V$ deformation retracts on $S^1$. Then,



$pi_1(mathbb{R}P^2)congmathbb{Z}*mathbb{Z}*_mathbb{Z} 1$. Let $i:Ucap Vto U$ be the inclusion, Then $pi_1(mathbb{R}P^2)conglangle a,bcolon i_*(1)rangle=langle a,bcolon ababrangle$










share|cite|improve this question











$endgroup$




I know that $pi_1(mathbb{R}P^2)congmathbb{Z}_2$, but in the square model, I get that $pi_1(mathbb{R}P^2)=langle a,bcolon ababrangle$. These groups must be isomorphic, but I can't find the isomorphism. What is the trick?



$mathbb{R}P^2$ is the following model



enter image description here



We want to use Seifert van Kampen, so let $U$ be the complement of a point and $V$ an open ball around that point. Then $U$ deformation retracts onto the boundary which is homotopy equivalent to $S^1vee S^1$. $V$ is simply connected and $Ucap V$ deformation retracts on $S^1$. Then,



$pi_1(mathbb{R}P^2)congmathbb{Z}*mathbb{Z}*_mathbb{Z} 1$. Let $i:Ucap Vto U$ be the inclusion, Then $pi_1(mathbb{R}P^2)conglangle a,bcolon i_*(1)rangle=langle a,bcolon ababrangle$







algebraic-topology projective-space fundamental-groups






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 11 at 11:23







James

















asked Mar 11 at 10:49









JamesJames

942318




942318












  • $begingroup$
    That's not the correct presentation; your second group maps onto $(mathbb{Z/2})^2$
    $endgroup$
    – Max
    Mar 11 at 11:08










  • $begingroup$
    Perhaps write down your "square model" and find the mistake.
    $endgroup$
    – Tyrone
    Mar 11 at 11:15










  • $begingroup$
    I have edited my question
    $endgroup$
    – James
    Mar 11 at 11:23










  • $begingroup$
    Is the mistake that $U$ doesn't deformation retract on the bouquet of circles?
    $endgroup$
    – James
    Mar 11 at 11:32


















  • $begingroup$
    That's not the correct presentation; your second group maps onto $(mathbb{Z/2})^2$
    $endgroup$
    – Max
    Mar 11 at 11:08










  • $begingroup$
    Perhaps write down your "square model" and find the mistake.
    $endgroup$
    – Tyrone
    Mar 11 at 11:15










  • $begingroup$
    I have edited my question
    $endgroup$
    – James
    Mar 11 at 11:23










  • $begingroup$
    Is the mistake that $U$ doesn't deformation retract on the bouquet of circles?
    $endgroup$
    – James
    Mar 11 at 11:32
















$begingroup$
That's not the correct presentation; your second group maps onto $(mathbb{Z/2})^2$
$endgroup$
– Max
Mar 11 at 11:08




$begingroup$
That's not the correct presentation; your second group maps onto $(mathbb{Z/2})^2$
$endgroup$
– Max
Mar 11 at 11:08












$begingroup$
Perhaps write down your "square model" and find the mistake.
$endgroup$
– Tyrone
Mar 11 at 11:15




$begingroup$
Perhaps write down your "square model" and find the mistake.
$endgroup$
– Tyrone
Mar 11 at 11:15












$begingroup$
I have edited my question
$endgroup$
– James
Mar 11 at 11:23




$begingroup$
I have edited my question
$endgroup$
– James
Mar 11 at 11:23












$begingroup$
Is the mistake that $U$ doesn't deformation retract on the bouquet of circles?
$endgroup$
– James
Mar 11 at 11:32




$begingroup$
Is the mistake that $U$ doesn't deformation retract on the bouquet of circles?
$endgroup$
– James
Mar 11 at 11:32










1 Answer
1






active

oldest

votes


















2












$begingroup$

First of all $U$ deformation retracts to the boundary which is homeomorphic to $mathbb Rmathbb P^1cong S^1$ and not $S^1lor S^1$. So by Van Kampen u get $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>$ where $omega$ is the generator of $pi_1(Ucap V)congpi_1(S^1)congmathbb Z$. But the generator deformation retracts to a path which winds twice around the boundary circle. Thus u get $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$






share|cite|improve this answer









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    $begingroup$

    First of all $U$ deformation retracts to the boundary which is homeomorphic to $mathbb Rmathbb P^1cong S^1$ and not $S^1lor S^1$. So by Van Kampen u get $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>$ where $omega$ is the generator of $pi_1(Ucap V)congpi_1(S^1)congmathbb Z$. But the generator deformation retracts to a path which winds twice around the boundary circle. Thus u get $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      First of all $U$ deformation retracts to the boundary which is homeomorphic to $mathbb Rmathbb P^1cong S^1$ and not $S^1lor S^1$. So by Van Kampen u get $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>$ where $omega$ is the generator of $pi_1(Ucap V)congpi_1(S^1)congmathbb Z$. But the generator deformation retracts to a path which winds twice around the boundary circle. Thus u get $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        First of all $U$ deformation retracts to the boundary which is homeomorphic to $mathbb Rmathbb P^1cong S^1$ and not $S^1lor S^1$. So by Van Kampen u get $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>$ where $omega$ is the generator of $pi_1(Ucap V)congpi_1(S^1)congmathbb Z$. But the generator deformation retracts to a path which winds twice around the boundary circle. Thus u get $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$






        share|cite|improve this answer









        $endgroup$



        First of all $U$ deformation retracts to the boundary which is homeomorphic to $mathbb Rmathbb P^1cong S^1$ and not $S^1lor S^1$. So by Van Kampen u get $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>$ where $omega$ is the generator of $pi_1(Ucap V)congpi_1(S^1)congmathbb Z$. But the generator deformation retracts to a path which winds twice around the boundary circle. Thus u get $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 11 at 11:36









        Soumik GhoshSoumik Ghosh

        760111




        760111






























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