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In an arcwise connected topological space $X$, we can show that the two groups $\pi(X,x)$ and $\pi(X,y)$ are isomorphic for $x,y \in X$ by defining a mapping $u: \pi(X,x) \to \pi(X,y)$ by $\alpha \mapsto \gamma^{-1} \alpha \gamma$. $\gamma$ is a path class with inital point $x$ and terminal point $y$. This is how William Massey does it in his book A Basic Course in Algebraic Topology.

My question is this: Why is $\gamma^{-1} \alpha \gamma$ an element of $\pi(X,y)$? As I see it, $\gamma$ takes $x$ to $y$, $\alpha$ makes a loop at $y$ and $\gamma^{-1}$ takes it back to $x$. So why is it an element of $\pi(X,y)$?

Hope you can help!

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  • $\begingroup$ In your notation $\alpha $ is a loop based at $x$- while in the body of the text you say that $\alpha $ is a loop based at $y$. Please check and let me know if I am missing something. $\endgroup$ – voldemort Aug 24 '14 at 1:28
  • $\begingroup$ @voldemort I understood $\gamma^{-1}\alpha \gamma$ as a way of transferring the loop to $y$ through $\gamma$ first. Does that make sense? $\alpha \in \pi(X,x)$, $\alpha \gamma \in \pi(X,y)$, $\gamma^{-1}\alpha \gamma \in \pi(X,x)$. There's something wrong with how I see things here. $\endgroup$ – Numbersandsoon Aug 24 '14 at 1:32
  • $\begingroup$ I must be missing something- I am still confused.. but I shall give up now :). It has been a long day. $\endgroup$ – voldemort Aug 24 '14 at 1:37
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    $\begingroup$ Your mistake is to use the wrong order in the concatenation of paths: the inverse of $\gamma$ comes first. This is not the same as the composition of functions. $\endgroup$ – Moishe Kohan Aug 24 '14 at 1:40
  • $\begingroup$ @studiosus Aaaah. I get it now. Haha, I almost want to apologize for haven gotten it the other way around :) It's so silly. Thanks studiosus. $\endgroup$ – Numbersandsoon Aug 24 '14 at 5:32
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I feel this situation is better understood as a basic fact on groupoids. Recall that a groupoid $G$ is a small category in which ever arrow is an isomorphism. Also $G$ is connected, or transitive, if for all objects $x,y$ of $G$, $G(x,y)$ is nonempty.

If $a \in G(x,y)$, and we write $G(x)$ for $G(x,x)$, then $a$ induces by conjugation an isomorphism $G(x) \to G(y)$, which is independent of the choice of $a$ in $G(x,y)$ if and only if $G(x)$ is abelian.

I don't write this out explicitly as there is a question of notation for composition: if $a \in G(x,y)$ and $b \in G(y,z)$ the their composite will lie in $G(x,z)$. Higgins book Categories and Groupoids writes it one way and my book Topology and Groupoids writes it the other!

The reasons for writing an abstract argument are as usual: 1) It applies to several known examples. 2) It will apply to new examples. 3) It simplifies the proof by concentrating on the relevant aspects.

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