Fundamental group of $mathbb{R}P^2$ in 2 modelsFundamental group of the torusGood exercises to do/examples to...
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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
$begingroup$
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
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
asked Mar 11 at 10:49
JamesJames
942318
$endgroup$
add a comment |
$begingroup$
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
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
asked Mar 11 at 10:49
JamesJames
942318
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
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
asked Mar 11 at 10:49
JamesJames
942318
$endgroup$
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
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
algebraic-topology projective-space fundamental-groups
asked Mar 11 at 10:49
JamesJames
942318
asked Mar 11 at 10:49
JamesJames
942318
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
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$
answered Mar 11 at 11:36
Soumik GhoshSoumik Ghosh
760111
$endgroup$
add a comment |
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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 $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$
answered Mar 11 at 11:36
Soumik GhoshSoumik Ghosh
760111
$endgroup$
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 $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$
answered Mar 11 at 11:36
Soumik GhoshSoumik Ghosh
760111
$endgroup$
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 $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$
answered Mar 11 at 11:36
Soumik GhoshSoumik Ghosh
760111
$endgroup$
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 $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$
answered Mar 11 at 11:36
Soumik GhoshSoumik Ghosh
760111
answered Mar 11 at 11:36
Soumik GhoshSoumik Ghosh
760111
answered Mar 11 at 11:36
Soumik GhoshSoumik Ghosh
760111
answered Mar 11 at 11:36
Soumik GhoshSoumik Ghosh
760111
760111
add a comment |
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$begingroup$
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