Fundamental groups in path-connected space

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The map $\psi$ is defined by $\psi(\alpha) = \gamma \alpha \gamma^{-1}$, where $\gamma:[0,1] \to X$ is a path with $\gamma(0) = y$, $\gamma(1) = x$. Then a homotopy can be given by "reeling in" $\gamma$.

A "reeling" homotopy is $$ h_s(t) = \begin{cases} \gamma(s) & t \in [0,s] \\ \gamma(t) & t \in [s,1]. \end{cases} $$ I'll leave you with the exercise to use this to get a homotopy from $\psi(\alpha)$ to $\alpha$.

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Irddo
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Irddo

Updated on May 21, 2020

Comments

  • Irddo
    Irddo over 3 years

    I'm studying Fundamental groups and today I saw the follow theorem:

    Theorem: Let be $X$ a topological space path-connected and $x,y\in X$. Then, the application $\psi:\pi_1(X,x)\to \pi_1(X,y)$ is a isomorphism of groups.

    I understood the proof, but I would like to know if, under conditions of the previous theorem, for each patch $\alpha\in\pi_1(X,x)$ there is a homotopy between $\alpha$ and $\psi(\alpha)$? I know this result is positive when $X$ is simply connected, but not in this case.

    Thank you!

    • Admin
      Admin over 8 years
      I assume you mean a free homotopy, considered as a map from $S^1$, yes? Because the question doesn't make sense if you want the homotopy to preserve basepoints (the two maps send the basepoint of $S^1$ to different points!) In this case, look at the construction of the map $\psi$ - the homotopy is visible in the construction.
    • Irddo
      Irddo over 8 years
      I'm sorry, and yes I'm talking of free homotopy. In the construction of $\psi$, basically I consider a patch between $y$ and $x$, is this order, denote by $\beta$, and for each $\alpha\in\pi_1(X,x)$ I define $\psi(\alpha)=\beta\alpha\beta^{-1}$, but I don't see why this solve my problem.
  • Irddo
    Irddo over 8 years
    Thank you for your answer, and I understood that is sufficiently "reeling in" $\gamma$ but can you to explicit the homotopy, please?
  • William Stagner
    William Stagner over 8 years
    Added a "reeling" homotopy