Cannonical group isomorpshim for fundamental group in a path connected space.












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 7:36







MUH

















asked Dec 1 '18 at 6:19









MUHMUH

410316




410316












  • $begingroup$
    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










  • $begingroup$
    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












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$begingroup$

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 ($textit{free homotopy}$).






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    $begingroup$

    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 ($textit{free homotopy}$).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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 ($textit{free homotopy}$).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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 ($textit{free homotopy}$).






        share|cite|improve this answer











        $endgroup$



        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 ($textit{free homotopy}$).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 '18 at 9:22

























        answered Dec 1 '18 at 9:12









        bourbaki22034bourbaki22034

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