Is repeating a loop twice the same as multiplication by $2$ in singular homology?












1












$begingroup$


I have the following problem understanding the notion of coefficients in Singular Homology. Let $X$ be a topological space (for example closed differential manifold). Let $x:[0,1]rightarrow X$ be a continous path such that $x(0)=x(1)$. Then $x$ defines a homology class $[x]in H_{1}(X;mathbb{Z})$. Define a different path $tilde{x}:[0,1]rightarrow X$ by $tilde{x}(t)=x(2t)$, for $tin [0,1/2]$ and $tilde{x}(t)=x(2t-1)$, for $tin [1/2,1]$ (essentially the loop $x$ run twice).



Question: Do we have that $2[x]=[tilde{x}]$ in $H_{1}(X;mathbb{Z})$?



If yes or no, then can you provide an explanation?



Greetings,
Milan






algebraic-topology homology-cohomology







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are basically using the Hurewicz homomorphism.
    $endgroup$
    – Justin Young
    Jan 7 at 17:19










  • $begingroup$
    Another formulation: as the map $tmapsto 2t$ is the double covering $S^1overset{times 2}{to} S^1$, the homology class $[tilde{x}]$ is the image of the fundamental class of $S^1$ under the map $S^1overset{times 2}{to} S^1overset{x}{to} X$. Thus it suffices to show that the double covering map doubles the fundamental clas of $S^1$, which you can find in most textbooks in algebraic topology, eg) Hatcher's.
    $endgroup$
    – cjackal
    Jan 7 at 21:57
















1












$begingroup$


I have the following problem understanding the notion of coefficients in Singular Homology. Let $X$ be a topological space (for example closed differential manifold). Let $x:[0,1]rightarrow X$ be a continous path such that $x(0)=x(1)$. Then $x$ defines a homology class $[x]in H_{1}(X;mathbb{Z})$. Define a different path $tilde{x}:[0,1]rightarrow X$ by $tilde{x}(t)=x(2t)$, for $tin [0,1/2]$ and $tilde{x}(t)=x(2t-1)$, for $tin [1/2,1]$ (essentially the loop $x$ run twice).



Question: Do we have that $2[x]=[tilde{x}]$ in $H_{1}(X;mathbb{Z})$?



If yes or no, then can you provide an explanation?



Greetings,
Milan






algebraic-topology homology-cohomology







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are basically using the Hurewicz homomorphism.
    $endgroup$
    – Justin Young
    Jan 7 at 17:19










  • $begingroup$
    Another formulation: as the map $tmapsto 2t$ is the double covering $S^1overset{times 2}{to} S^1$, the homology class $[tilde{x}]$ is the image of the fundamental class of $S^1$ under the map $S^1overset{times 2}{to} S^1overset{x}{to} X$. Thus it suffices to show that the double covering map doubles the fundamental clas of $S^1$, which you can find in most textbooks in algebraic topology, eg) Hatcher's.
    $endgroup$
    – cjackal
    Jan 7 at 21:57














1












1








1





$begingroup$


I have the following problem understanding the notion of coefficients in Singular Homology. Let $X$ be a topological space (for example closed differential manifold). Let $x:[0,1]rightarrow X$ be a continous path such that $x(0)=x(1)$. Then $x$ defines a homology class $[x]in H_{1}(X;mathbb{Z})$. Define a different path $tilde{x}:[0,1]rightarrow X$ by $tilde{x}(t)=x(2t)$, for $tin [0,1/2]$ and $tilde{x}(t)=x(2t-1)$, for $tin [1/2,1]$ (essentially the loop $x$ run twice).



Question: Do we have that $2[x]=[tilde{x}]$ in $H_{1}(X;mathbb{Z})$?



If yes or no, then can you provide an explanation?



Greetings,
Milan






algebraic-topology homology-cohomology







share|cite|improve this question











$endgroup$




I have the following problem understanding the notion of coefficients in Singular Homology. Let $X$ be a topological space (for example closed differential manifold). Let $x:[0,1]rightarrow X$ be a continous path such that $x(0)=x(1)$. Then $x$ defines a homology class $[x]in H_{1}(X;mathbb{Z})$. Define a different path $tilde{x}:[0,1]rightarrow X$ by $tilde{x}(t)=x(2t)$, for $tin [0,1/2]$ and $tilde{x}(t)=x(2t-1)$, for $tin [1/2,1]$ (essentially the loop $x$ run twice).



Question: Do we have that $2[x]=[tilde{x}]$ in $H_{1}(X;mathbb{Z})$?



If yes or no, then can you provide an explanation?



Greetings,
Milan






algebraic-topology homology-cohomology




algebraic-topology homology-cohomology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 21:45









Eric Wofsey

182k13209337




182k13209337










asked Jan 7 at 16:48









MilanMilan

83




83












  • $begingroup$
    You are basically using the Hurewicz homomorphism.
    $endgroup$
    – Justin Young
    Jan 7 at 17:19










  • $begingroup$
    Another formulation: as the map $tmapsto 2t$ is the double covering $S^1overset{times 2}{to} S^1$, the homology class $[tilde{x}]$ is the image of the fundamental class of $S^1$ under the map $S^1overset{times 2}{to} S^1overset{x}{to} X$. Thus it suffices to show that the double covering map doubles the fundamental clas of $S^1$, which you can find in most textbooks in algebraic topology, eg) Hatcher's.
    $endgroup$
    – cjackal
    Jan 7 at 21:57


















  • $begingroup$
    You are basically using the Hurewicz homomorphism.
    $endgroup$
    – Justin Young
    Jan 7 at 17:19










  • $begingroup$
    Another formulation: as the map $tmapsto 2t$ is the double covering $S^1overset{times 2}{to} S^1$, the homology class $[tilde{x}]$ is the image of the fundamental class of $S^1$ under the map $S^1overset{times 2}{to} S^1overset{x}{to} X$. Thus it suffices to show that the double covering map doubles the fundamental clas of $S^1$, which you can find in most textbooks in algebraic topology, eg) Hatcher's.
    $endgroup$
    – cjackal
    Jan 7 at 21:57
















$begingroup$
You are basically using the Hurewicz homomorphism.
$endgroup$
– Justin Young
Jan 7 at 17:19




$begingroup$
You are basically using the Hurewicz homomorphism.
$endgroup$
– Justin Young
Jan 7 at 17:19












$begingroup$
Another formulation: as the map $tmapsto 2t$ is the double covering $S^1overset{times 2}{to} S^1$, the homology class $[tilde{x}]$ is the image of the fundamental class of $S^1$ under the map $S^1overset{times 2}{to} S^1overset{x}{to} X$. Thus it suffices to show that the double covering map doubles the fundamental clas of $S^1$, which you can find in most textbooks in algebraic topology, eg) Hatcher's.
$endgroup$
– cjackal
Jan 7 at 21:57




$begingroup$
Another formulation: as the map $tmapsto 2t$ is the double covering $S^1overset{times 2}{to} S^1$, the homology class $[tilde{x}]$ is the image of the fundamental class of $S^1$ under the map $S^1overset{times 2}{to} S^1overset{x}{to} X$. Thus it suffices to show that the double covering map doubles the fundamental clas of $S^1$, which you can find in most textbooks in algebraic topology, eg) Hatcher's.
$endgroup$
– cjackal
Jan 7 at 21:57










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Yes. Consider an oriented equilateral triangle $Delta=e_0e_1e_2$, and let $f(P)=P’$ for every $P in Delta$, where $P’ in [e_0,e_2]$ and the lines $PP’$ and $e_0e_2$ are perpendicular.



Then the boundary of $tilde{x} circ f$ is exactly $x+x-tilde{x}$.






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    $begingroup$

    Yes. Consider an oriented equilateral triangle $Delta=e_0e_1e_2$, and let $f(P)=P’$ for every $P in Delta$, where $P’ in [e_0,e_2]$ and the lines $PP’$ and $e_0e_2$ are perpendicular.



    Then the boundary of $tilde{x} circ f$ is exactly $x+x-tilde{x}$.






    share|cite|improve this answer









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      2












      $begingroup$

      Yes. Consider an oriented equilateral triangle $Delta=e_0e_1e_2$, and let $f(P)=P’$ for every $P in Delta$, where $P’ in [e_0,e_2]$ and the lines $PP’$ and $e_0e_2$ are perpendicular.



      Then the boundary of $tilde{x} circ f$ is exactly $x+x-tilde{x}$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Yes. Consider an oriented equilateral triangle $Delta=e_0e_1e_2$, and let $f(P)=P’$ for every $P in Delta$, where $P’ in [e_0,e_2]$ and the lines $PP’$ and $e_0e_2$ are perpendicular.



        Then the boundary of $tilde{x} circ f$ is exactly $x+x-tilde{x}$.






        share|cite|improve this answer









        $endgroup$



        Yes. Consider an oriented equilateral triangle $Delta=e_0e_1e_2$, and let $f(P)=P’$ for every $P in Delta$, where $P’ in [e_0,e_2]$ and the lines $PP’$ and $e_0e_2$ are perpendicular.



        Then the boundary of $tilde{x} circ f$ is exactly $x+x-tilde{x}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 17:06









        MindlackMindlack

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