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Fundamental group of $mathbbRP^2$ 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 $pi_1(mathbbRP^2)congmathbbZ_2$, but in the square model, I get that $pi_1(mathbbRP^2)=langle a,bcolon ababrangle$. These groups must be isomorphic, but I can't find the isomorphism. What is the trick?

$mathbbRP^2$ is the following model

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,

$pi_1(mathbbRP^2)congmathbbZ*mathbbZ*_mathbbZ 1$. Let $i:Ucap Vto U$ be the inclusion, Then $pi_1(mathbbRP^2)conglangle a,bcolon i_*(1)rangle=langle a,bcolon ababrangle$

algebraic-topology projective-space fundamental-groups

share|cite|improve this question

edited Mar 11 at 11:23

James

asked Mar 11 at 10:49

JamesJames

942318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's not the correct presentation; your second group maps onto $(mathbbZ/2)^2$
  $endgroup$
  – 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
  

  
  
  
  

add a comment | 

1

$begingroup$

I know that $pi_1(mathbbRP^2)congmathbbZ_2$, but in the square model, I get that $pi_1(mathbbRP^2)=langle a,bcolon ababrangle$. These groups must be isomorphic, but I can't find the isomorphism. What is the trick?

$mathbbRP^2$ is the following model

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,

$pi_1(mathbbRP^2)congmathbbZ*mathbbZ*_mathbbZ 1$. Let $i:Ucap Vto U$ be the inclusion, Then $pi_1(mathbbRP^2)conglangle a,bcolon i_*(1)rangle=langle a,bcolon ababrangle$

algebraic-topology projective-space fundamental-groups

share|cite|improve this question

edited Mar 11 at 11:23

James

asked Mar 11 at 10:49

JamesJames

942318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's not the correct presentation; your second group maps onto $(mathbbZ/2)^2$
  $endgroup$
  – 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
  

  
  
  
  

add a comment | 

$begingroup$

I know that $pi_1(mathbbRP^2)congmathbbZ_2$, but in the square model, I get that $pi_1(mathbbRP^2)=langle a,bcolon ababrangle$. These groups must be isomorphic, but I can't find the isomorphism. What is the trick?

$mathbbRP^2$ is the following model

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,

$pi_1(mathbbRP^2)congmathbbZ*mathbbZ*_mathbbZ 1$. Let $i:Ucap Vto U$ be the inclusion, Then $pi_1(mathbbRP^2)conglangle a,bcolon i_*(1)rangle=langle a,bcolon ababrangle$

algebraic-topology projective-space fundamental-groups

share|cite|improve this question

edited Mar 11 at 11:23

James

asked Mar 11 at 10:49

JamesJames

942318

$endgroup$

share|cite|improve this question

edited Mar 11 at 11:23

James

asked Mar 11 at 10:49

JamesJames

942318

share|cite|improve this question

edited Mar 11 at 11:23

James

asked Mar 11 at 10:49

JamesJames

942318

asked Mar 11 at 10:49

JamesJames

942318

asked Mar 11 at 10:49

JamesJames

942318

1 Answer
1

active

oldest

votes

1

$begingroup$

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

answered Mar 11 at 11:36

Soumik GhoshSoumik Ghosh

675111

$endgroup$

add a comment | 

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1

$begingroup$

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

answered Mar 11 at 11:36

Soumik GhoshSoumik Ghosh

675111

$endgroup$

add a comment | 

1

$begingroup$

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

answered Mar 11 at 11:36

Soumik GhoshSoumik Ghosh

675111

$endgroup$

add a comment | 

$begingroup$

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

answered Mar 11 at 11:36

Soumik GhoshSoumik Ghosh

675111

$endgroup$

share|cite|improve this answer

answered Mar 11 at 11:36

Soumik GhoshSoumik Ghosh

675111

answered Mar 11 at 11:36

Soumik GhoshSoumik Ghosh

675111

answered Mar 11 at 11:36

Soumik GhoshSoumik Ghosh

675111