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Let $\overline{X}$ be a universal covering space of $X$. If $p:\overline{X}\to X$ is the projection, then let $p^{-1}(x)$ be the fiber of $x\in X$, and let $q\in p^{-1}(x)$. Consider the action of the fundamental group $\pi_1(X)$ on $q$ in two different ways:

  1. For any $[c]\in\pi_1(X)$, consider the action of the lift on $q$.
  2. Consider the action of the Deck transformation corresponding to $[c]$ on $q$.

Are these two actions ever different? I think both actions give the end point of the lift of $[c]$ starting at $q$. However, my professor says that these two actions are different when $X$ is the wedged circle. I don't quite understand this statement. Any help would be great.

  • What is your definition of the deck transformation corresponding to $[c]$? – Eric Wofsey Mar 29 '16 at 07:27
  • @EricWofsey- I am not clear on the definition. What I understand: let $d$ be the Deck transformation corresponding to $[c]$ (such a correspondence exists as the group of deck transformations and the fundamental group are isomorphic. Then $d.q=c.q= $ the end point of the lift of $[c]$. –  Mar 29 '16 at 07:34
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    I've never heard of "the wedged circle". If you have a universal cover, then the fundamental group is naturally anti-isomorphic to the group of automorphisms of the cover ("deck transformations"), and the (left) tautological action of the automorphism group and the (right) monodromy action of the fundamental group are conjugated by this anti-automorphism. I don't really know what your professor is referring to. – Captain Lama Mar 29 '16 at 09:08
  • Perhaps "the wedged circle" is a rewording of "the wedge of two circles", in which case your comment is really an answer. @CaptainLama – Lee Mosher Mar 29 '16 at 11:52
  • @CaptainLama- Could you please expand this as an answer? I don't know what anti-isomorphic is, for instance. –  Mar 29 '16 at 15:25
  • The two actions agree if for $M \xrightarrow{\pi} N$ the group $\pi_1(N)$ is abelian. So look for counterexamples there. The Deck group operates on the fibers via the following isomorphism: take the stabelizer of a point in the fiber $\Pi$ and then consider the $N(\Pi) \to \operatorname{Deck}$. In Bredon's "Topology and Geometry" should be details. – CPJ Dec 02 '23 at 09:10

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