Fundamental group of $mathbbRP^2$ in 2 modelsFundamental 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
What is IP squat space
Sword in the Stone story where the sword was held in place by electromagnets
Could the Saturn V actually have launched astronauts around Venus?
What are some nice/clever ways to introduce the tonic's dominant seventh chord?
What are the possible solutions of the given equation?
How to generate globally unique ids for different tables of the same database?
How to deal with taxi scam when on vacation?
How could a scammer know the apps on my phone / iTunes account?
Meaning of "SEVERA INDEOVI VAS" from 3rd Century slab
What options are left, if Britain cannot decide?
Why do passenger jet manufacturers design their planes with stall prevention systems?
Why doesn't the EU now just force the UK to choose between referendum and no-deal?
Can hydraulic brake levers get hot when brakes overheat?
How to make healing in an exploration game interesting
Is it normal that my co-workers at a fitness company criticize my food choices?
How do anti-virus programs start at Windows boot?
Use of プラトニック in this sentence?
Co-worker team leader wants to inject his friend's awful software into our development. What should I say to our common boss?
Current sense amp + op-amp buffer + ADC: Measuring down to 0 with single supply
Identifying the interval from A♭ to D♯
How can I change step-down my variable input voltage? [Microcontroller]
Can elves maintain concentration in a trance?
How to write cleanly even if my character uses expletive language?
Science-fiction short story where space navy wanted hospital ships and settlers had guns mounted everywhere
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
$begingroup$
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
asked Mar 11 at 10:49
JamesJames
942318
$endgroup$
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
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
asked Mar 11 at 10:49
JamesJames
942318
$endgroup$
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
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 $(mathbbZ/2)^2$
$endgroup$
– 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 |
$begingroup$
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
$begingroup$
That's not the correct presentation; your second group maps onto $(mathbbZ/2)^2$
$endgroup$
– Max
Mar 11 at 11:08
$begingroup$
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
$begingroup$
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
675111
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3143552%2ffundamental-group-of-mathbbrp2-in-2-models%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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
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 $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$
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 $pi_1(mathbb R mathbb P^1)=mathbb Z/<i_*(omega)>=<a|a^2>congmathbb Z_2$
answered Mar 11 at 11:36
Soumik GhoshSoumik Ghosh
675111
$endgroup$
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
675111
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
675111
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3143552%2ffundamental-group-of-mathbbrp2-in-2-models%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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