Cannonical group isomorpshim for fundamental group in a path connected space.
$begingroup$
My question is related to this one which tells us that fundamental group $pi_{1}(X,x_0)$ is abelian $textit{iff}$ for every pair $alpha$ and $beta$ of paths from $x_0$ to $x_1$, we have the same group isomorphism.
let $[I,X]$ denote the set of homotopy classes of maps of $I$ into $X$, where $I=[0,1]$. If $X$ is path connected, we can easily prove that $[I,X]$ has only one element, which mean that every path in $X$ is homotopic to each other (even a loop at $x_0 in X$ is homotopic to path $p:x_0 rightarrow x_1$).
using this fact,
$hat{alpha}([f]) = [bar{alpha}]*[f]*[alpha] = [bar{beta}]*[f]*[beta]=hat{beta}([f]) $, wehre $alpha$ and $beta$ denote paths in path connected space $X$ ($alpha simeq beta Rightarrow [alpha]= [beta] $ ). So, it seems that there is no need for the fundamental group $pi_{1}(X,x_0)$ to be abelian.
Also, for the proof that $pi_{1}(X,x_0)$ is abelian, if you take 2 loops at $x_0$, then $[f]*[g]=[g]*[f]$ by virtue of being two paths in path-connected space $X$.
Am I wrong here? Where am I wrong?
algebraic-topology fundamental-groups
asked Dec 1 '18 at 6:19
MUHMUH
395216
$endgroup$
add a comment |
$begingroup$
My question is related to this one which tells us that fundamental group $pi_{1}(X,x_0)$ is abelian $textit{iff}$ for every pair $alpha$ and $beta$ of paths from $x_0$ to $x_1$, we have the same group isomorphism.
let $[I,X]$ denote the set of homotopy classes of maps of $I$ into $X$, where $I=[0,1]$. If $X$ is path connected, we can easily prove that $[I,X]$ has only one element, which mean that every path in $X$ is homotopic to each other (even a loop at $x_0 in X$ is homotopic to path $p:x_0 rightarrow x_1$).
using this fact,
$hat{alpha}([f]) = [bar{alpha}]*[f]*[alpha] = [bar{beta}]*[f]*[beta]=hat{beta}([f]) $, wehre $alpha$ and $beta$ denote paths in path connected space $X$ ($alpha simeq beta Rightarrow [alpha]= [beta] $ ). So, it seems that there is no need for the fundamental group $pi_{1}(X,x_0)$ to be abelian.
Also, for the proof that $pi_{1}(X,x_0)$ is abelian, if you take 2 loops at $x_0$, then $[f]*[g]=[g]*[f]$ by virtue of being two paths in path-connected space $X$.
Am I wrong here? Where am I wrong?
algebraic-topology fundamental-groups
asked Dec 1 '18 at 6:19
MUHMUH
395216
$endgroup$
$begingroup$
You are confusing free homotopies with base-point preserving homotopies.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 6:48
$begingroup$
I read this, but I am not sure how this concept of free homotopies and base-point preserving homotopies answer my question. One of the points that I still can't find any fault is that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq beta $(since $X$ is path connected) which imply that $ [alpha] = [beta]$ and hence, $hat{alpha}= hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:29
$begingroup$
@LordSharktheUnknown Also, if the free homotopies hold true, and given the base-point maps, (for e.g. here $hat{alpha},hat{beta}$, which are base-point maps at $x_1$ or $textit{loops}$ at $x_1$), doesn't it imply $hat{alpha}=hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:40
add a comment |
$begingroup$
My question is related to this one which tells us that fundamental group $pi_{1}(X,x_0)$ is abelian $textit{iff}$ for every pair $alpha$ and $beta$ of paths from $x_0$ to $x_1$, we have the same group isomorphism.
let $[I,X]$ denote the set of homotopy classes of maps of $I$ into $X$, where $I=[0,1]$. If $X$ is path connected, we can easily prove that $[I,X]$ has only one element, which mean that every path in $X$ is homotopic to each other (even a loop at $x_0 in X$ is homotopic to path $p:x_0 rightarrow x_1$).
using this fact,
$hat{alpha}([f]) = [bar{alpha}]*[f]*[alpha] = [bar{beta}]*[f]*[beta]=hat{beta}([f]) $, wehre $alpha$ and $beta$ denote paths in path connected space $X$ ($alpha simeq beta Rightarrow [alpha]= [beta] $ ). So, it seems that there is no need for the fundamental group $pi_{1}(X,x_0)$ to be abelian.
Also, for the proof that $pi_{1}(X,x_0)$ is abelian, if you take 2 loops at $x_0$, then $[f]*[g]=[g]*[f]$ by virtue of being two paths in path-connected space $X$.
Am I wrong here? Where am I wrong?
algebraic-topology fundamental-groups
asked Dec 1 '18 at 6:19
MUHMUH
395216
$endgroup$
My question is related to this one which tells us that fundamental group $pi_{1}(X,x_0)$ is abelian $textit{iff}$ for every pair $alpha$ and $beta$ of paths from $x_0$ to $x_1$, we have the same group isomorphism.
let $[I,X]$ denote the set of homotopy classes of maps of $I$ into $X$, where $I=[0,1]$. If $X$ is path connected, we can easily prove that $[I,X]$ has only one element, which mean that every path in $X$ is homotopic to each other (even a loop at $x_0 in X$ is homotopic to path $p:x_0 rightarrow x_1$).
using this fact,
$hat{alpha}([f]) = [bar{alpha}]*[f]*[alpha] = [bar{beta}]*[f]*[beta]=hat{beta}([f]) $, wehre $alpha$ and $beta$ denote paths in path connected space $X$ ($alpha simeq beta Rightarrow [alpha]= [beta] $ ). So, it seems that there is no need for the fundamental group $pi_{1}(X,x_0)$ to be abelian.
Also, for the proof that $pi_{1}(X,x_0)$ is abelian, if you take 2 loops at $x_0$, then $[f]*[g]=[g]*[f]$ by virtue of being two paths in path-connected space $X$.
Am I wrong here? Where am I wrong?
algebraic-topology fundamental-groups
algebraic-topology fundamental-groups
asked Dec 1 '18 at 6:19
MUHMUH
395216
asked Dec 1 '18 at 6:19
MUHMUH
395216
edited Dec 1 '18 at 7:36
MUH
asked Dec 1 '18 at 6:19
MUHMUH
395216
asked Dec 1 '18 at 6:19
MUHMUH
395216
asked Dec 1 '18 at 6:19
MUHMUH
395216
395216
$begingroup$
You are confusing free homotopies with base-point preserving homotopies.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 6:48
$begingroup$
I read this, but I am not sure how this concept of free homotopies and base-point preserving homotopies answer my question. One of the points that I still can't find any fault is that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq beta $(since $X$ is path connected) which imply that $ [alpha] = [beta]$ and hence, $hat{alpha}= hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:29
$begingroup$
@LordSharktheUnknown Also, if the free homotopies hold true, and given the base-point maps, (for e.g. here $hat{alpha},hat{beta}$, which are base-point maps at $x_1$ or $textit{loops}$ at $x_1$), doesn't it imply $hat{alpha}=hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:40
add a comment |
$begingroup$
You are confusing free homotopies with base-point preserving homotopies.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 6:48
$begingroup$
I read this, but I am not sure how this concept of free homotopies and base-point preserving homotopies answer my question. One of the points that I still can't find any fault is that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq beta $(since $X$ is path connected) which imply that $ [alpha] = [beta]$ and hence, $hat{alpha}= hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:29
$begingroup$
@LordSharktheUnknown Also, if the free homotopies hold true, and given the base-point maps, (for e.g. here $hat{alpha},hat{beta}$, which are base-point maps at $x_1$ or $textit{loops}$ at $x_1$), doesn't it imply $hat{alpha}=hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:40
$begingroup$
You are confusing free homotopies with base-point preserving homotopies.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 6:48
$begingroup$
You are confusing free homotopies with base-point preserving homotopies.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 6:48
$begingroup$
I read this, but I am not sure how this concept of free homotopies and base-point preserving homotopies answer my question. One of the points that I still can't find any fault is that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq beta $(since $X$ is path connected) which imply that $ [alpha] = [beta]$ and hence, $hat{alpha}= hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:29
$begingroup$
I read this, but I am not sure how this concept of free homotopies and base-point preserving homotopies answer my question. One of the points that I still can't find any fault is that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq beta $(since $X$ is path connected) which imply that $ [alpha] = [beta]$ and hence, $hat{alpha}= hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:29
$begingroup$
@LordSharktheUnknown Also, if the free homotopies hold true, and given the base-point maps, (for e.g. here $hat{alpha},hat{beta}$, which are base-point maps at $x_1$ or $textit{loops}$ at $x_1$), doesn't it imply $hat{alpha}=hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:40
$begingroup$
@LordSharktheUnknown Also, if the free homotopies hold true, and given the base-point maps, (for e.g. here $hat{alpha},hat{beta}$, which are base-point maps at $x_1$ or $textit{loops}$ at $x_1$), doesn't it imply $hat{alpha}=hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your conclusion that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq_p beta$(since $X$ is path-connected) is wrong. (I think you mean path homotopy). It is true that any two paths in path connected space is homotpic to each other but not path-homotopy.
$p:x_0 rightarrow x_1$ and $q:y_0 rightarrow y_1$ are any two paths in path-connected space $X$. Then $p simeq e_{x_0}$ and $q simeq e_{y_0} $ and since, $e_{x_0} simeq e_{y_{0}} Rightarrow p simeq q $. But this doesn't mean that $p simeq_p q$. There is a difference between path-homotopy and homotopy that path-homotopy has to satisfy additional criteria of initial point and final point for each $t in I$, which clearly doesn't get satisfied in the proof of homotopy ($textit{free homotopy}$).
answered Dec 1 '18 at 9:12
bourbaki22034bourbaki22034
614
$endgroup$
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%2f3021053%2fcannonical-group-isomorpshim-for-fundamental-group-in-a-path-connected-space%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
$begingroup$
Your conclusion that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq_p beta$(since $X$ is path-connected) is wrong. (I think you mean path homotopy). It is true that any two paths in path connected space is homotpic to each other but not path-homotopy.
$p:x_0 rightarrow x_1$ and $q:y_0 rightarrow y_1$ are any two paths in path-connected space $X$. Then $p simeq e_{x_0}$ and $q simeq e_{y_0} $ and since, $e_{x_0} simeq e_{y_{0}} Rightarrow p simeq q $. But this doesn't mean that $p simeq_p q$. There is a difference between path-homotopy and homotopy that path-homotopy has to satisfy additional criteria of initial point and final point for each $t in I$, which clearly doesn't get satisfied in the proof of homotopy ($textit{free homotopy}$).
answered Dec 1 '18 at 9:12
bourbaki22034bourbaki22034
614
$endgroup$
add a comment |
$begingroup$
Your conclusion that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq_p beta$(since $X$ is path-connected) is wrong. (I think you mean path homotopy). It is true that any two paths in path connected space is homotpic to each other but not path-homotopy.
$p:x_0 rightarrow x_1$ and $q:y_0 rightarrow y_1$ are any two paths in path-connected space $X$. Then $p simeq e_{x_0}$ and $q simeq e_{y_0} $ and since, $e_{x_0} simeq e_{y_{0}} Rightarrow p simeq q $. But this doesn't mean that $p simeq_p q$. There is a difference between path-homotopy and homotopy that path-homotopy has to satisfy additional criteria of initial point and final point for each $t in I$, which clearly doesn't get satisfied in the proof of homotopy ($textit{free homotopy}$).
answered Dec 1 '18 at 9:12
bourbaki22034bourbaki22034
614
$endgroup$
add a comment |
$begingroup$
Your conclusion that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq_p beta$(since $X$ is path-connected) is wrong. (I think you mean path homotopy). It is true that any two paths in path connected space is homotpic to each other but not path-homotopy.
$p:x_0 rightarrow x_1$ and $q:y_0 rightarrow y_1$ are any two paths in path-connected space $X$. Then $p simeq e_{x_0}$ and $q simeq e_{y_0} $ and since, $e_{x_0} simeq e_{y_{0}} Rightarrow p simeq q $. But this doesn't mean that $p simeq_p q$. There is a difference between path-homotopy and homotopy that path-homotopy has to satisfy additional criteria of initial point and final point for each $t in I$, which clearly doesn't get satisfied in the proof of homotopy ($textit{free homotopy}$).
answered Dec 1 '18 at 9:12
bourbaki22034bourbaki22034
614
$endgroup$
Your conclusion that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq_p beta$(since $X$ is path-connected) is wrong. (I think you mean path homotopy). It is true that any two paths in path connected space is homotpic to each other but not path-homotopy.
$p:x_0 rightarrow x_1$ and $q:y_0 rightarrow y_1$ are any two paths in path-connected space $X$. Then $p simeq e_{x_0}$ and $q simeq e_{y_0} $ and since, $e_{x_0} simeq e_{y_{0}} Rightarrow p simeq q $. But this doesn't mean that $p simeq_p q$. There is a difference between path-homotopy and homotopy that path-homotopy has to satisfy additional criteria of initial point and final point for each $t in I$, which clearly doesn't get satisfied in the proof of homotopy ($textit{free homotopy}$).
answered Dec 1 '18 at 9:12
bourbaki22034bourbaki22034
614
edited Dec 1 '18 at 9:22
answered Dec 1 '18 at 9:12
bourbaki22034bourbaki22034
614
answered Dec 1 '18 at 9:12
bourbaki22034bourbaki22034
614
answered Dec 1 '18 at 9:12
bourbaki22034bourbaki22034
614
614
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%2f3021053%2fcannonical-group-isomorpshim-for-fundamental-group-in-a-path-connected-space%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
$begingroup$
You are confusing free homotopies with base-point preserving homotopies.
$endgroup$
– Lord Shark the Unknown
Dec 1 '18 at 6:48
$begingroup$
I read this, but I am not sure how this concept of free homotopies and base-point preserving homotopies answer my question. One of the points that I still can't find any fault is that if $alpha, beta : x_0 rightarrow x_1$, then $alpha simeq beta $(since $X$ is path connected) which imply that $ [alpha] = [beta]$ and hence, $hat{alpha}= hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:29
$begingroup$
@LordSharktheUnknown Also, if the free homotopies hold true, and given the base-point maps, (for e.g. here $hat{alpha},hat{beta}$, which are base-point maps at $x_1$ or $textit{loops}$ at $x_1$), doesn't it imply $hat{alpha}=hat{beta}$
$endgroup$
– MUH
Dec 1 '18 at 8:40