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In exercise 1.3.28 in Hatcher's Algebraic Topology, we are asked to show that for a covering space action of a group $G$ on a simply-connected space $Y$, $\pi_1(Y/G)$ is isomorphic to $G$. This is a special case of Proposition 1.40(c).

My question is: What happens if $Y$ is not locally path-connected and not simply connected? Is there a counterexample to show that Proposition 1.40(c) fails? I am still assuming $Y$ is path-connected.

Edit: I should clarify that I want to find a counterexample to proposition 1.40(c) if $Y$ is not locally path-connected, which means that I want to find a a covering space action of a group $G$ on a path-connected space $Y$ such that $G \ncong \pi_1(Y/G)/p_*(\pi_1(Y))$, where $p: Y \to Y/G$ is the quotient map. Proposition 1.40(c) implies that such a $Y$ cannot be locally path-connected, while Exercise 1.3.28 tells us that such a $Y$ cannot be simply connected.

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$S^1$ is the quotient of $R$ by $f(x)=x+1$, $Z/2$ acts on $S^1$ by the action induced by the translation $g(x)=x+1/2$ and the quotient of this action is still $S^1$.

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  • $\begingroup$ Sorry, but I don't get this counterexample. What are your $Y$ and $G$ in this case? $\endgroup$
    – Math Maniac
    Dec 30, 2015 at 17:26
  • $\begingroup$ $Y$ is $S^1$ and $G$ is $Z/2$ $\endgroup$
    – Tsemo Aristide
    Dec 30, 2015 at 17:27
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    $\begingroup$ How can this be a counterexample though - here $Y$ is path-connected and locally path-connected so cannot serve as a counterexample to Proposition 1.40(c). I want a counterexample to the proposition, not to the exercise, if $Y$ fails to satisfy the hypothesis of being locally path-connected. $\endgroup$
    – Math Maniac
    Dec 30, 2015 at 17:40

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