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We already know the theorem


Let $p: (Y,y) \rightarrow (X,x)$ be a covering, with $Y$ connected and $X$ locally path connected, and let $p(y) = x$. If $p_*(\pi_1(Y,y))$ is a normal subgroup of $\pi_1(X,x)$, then $\pi_1(X,x)/p_*(\pi_1(Y,y))$ is isomorphic to Aut(Y/X).


As the situation above, T.F.A.E

(a) the covering is normal (i.e., $p_*(\pi_1(Y,y))$ is a normal subgroup of $\pi_1(X,x)$);

(b) the action of $Aut(X/Y)$ on $p^{-1}(x)$ is transitive;

(c) for every loop $\sigma$ at $x$, if one lifting of $\sigma$ is closed (i.e., still a loop at some $y \in p^{-1}(x)$), then all liftings are closed.

My approach

I already proved (b) $\implies$ (c) and (c) $\implies$ (a). So I need to prove (a) $\implies$ (b), and I know I need to use the theorem above.

So, is that true that for any $y$ and $y' \in p^{-1}(x)$, the there exists $[\sigma] \in \pi_1(X,x)$ such that $y' = [\sigma]y$? (If this is true, then we are done!)

Thank you very much!

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up vote 1 down vote accepted

Yes, your property is true : $\pi_1(X,x)$ acts transitively on $p^{-1}(x)$ :

First, assume that $Y$ is path connected : let $y,y'\in p^{-1}(x)$, you can find a path $\gamma:y\rightsquigarrow y'$. Then notice that $\gamma$ is the unique lift of $p\circ\gamma:x\rightsquigarrow x$ with starting point y. So $y'=\gamma(1)=y\cdot[p\circ\gamma]$.

Now, notice that $p$ is a local homeomorphism, so $Y$ is connected and locally path connected (since $X$ is), so $Y$ is path connected.

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Thank you very much! I can understand your answer! – Peter Hu Jun 16 '12 at 12:37

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