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Suppose $X$ a free $G$-complex (i.e. a CW-complex with a free $G$-action that permutes the cells). I would like to show that the projection $$X\overset{p}{\to}X/G$$ is a regular covering spaces with covering transformations $Aut(p)\cong G$. This is an exercise from Brown's book "Cohomology of groups", and the first step, as indicated, should be proving that for every point $x\in X$ there exists a neighbourhood $U_x$ such that $gU_x\cap U_x=\emptyset$ for every $g\neq 1$. Now, this should be quite easy, and it should follow directly from the fact that the action is free and that no two simplices can intersect if not in a face. Now, this should imply that $X\to X/G$ is a covering: for every $y\in X/G$ I take a point in the counterimage, say $x$, and I choose $U_x$ as before. Now, it is clear that $G$ maps $U_x$ to other neighbourhoods of the points in the counterimage and that these neighbourhoods satisfy the same property $gU\cap U=\emptyset$. Assuming that this is correct (up to details!), how can I conclude that $Aut(p)\cong G$? Thank you in advance, bye!

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You need to suppose that $X$ is connected (otherwise it's not true: see e.g. $X=$ $n$ points, $G=$ cyclic of order $n$, $Aut=S_n$). Certainly $G\subset Aut(p)$. Let $x_0\in X$. Any automorphism $a$ of $X\to X/G$ is uniquely determined if we know where $a$ maps $x_0$ (here we need to know that $X$ is connected). And $a(x_0)=g\cdot x_0$ for some $g\in G$ (as $p(a(x_0))=p(x_0)$), hence $a=g$.

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