Basic question about G-coverings

Question

Given a topological space $X$, and a group $G$ acting on $X$, I don't understand why the projection $\pi:X\to X/G$ isn't a covering map if the action of $G$ has fixed points. In other words I don't understand why if, given a covering map $p:E \to Y$, the action of $\mathrm{Aut}(E,p)$ must not have fixed points in order to have $Y\simeq E/\mathrm{Aut}(E,p)$. It's clear to me that if the action of $G=\mathrm{Aut}(E,p)$ has no fixed points then $E\simeq Y\times G$; however even if the action has some fixed points, we can glue every two points $e_1, e_2\in E$ if there is some $g\in G$ such that $e_1=ge_2$. After this process we obtain the space $Y$, and this should mean that $Y\simeq E/G$. Where am I wrong? Maybe we can't glue the points without "ripping" the space $E$? Thank you very much

Answer 1

"It's clear to me that if the action has no fixed point then $E\simeq Y\times G$ " : that's really far from true; if it were, coverings would be really uninteresting.

Think about $\mathbb Z$ acting on $\mathbb R$ by translation for instance.

The problem is that, although $E/G$ still exists the projection map $E\to E/G$ is not a covering in general (it can be : if $G$ acts trivially, it has tons of fixed points and $E=E/G$ so $E\to E/G$ is a covering map).

To understand why that is, consider $\mathbb Z/2$ acting on $\mathbb R$ by $-1$ : $0$ is a fixed point, and the quotient is homeomorphic to $\mathbb R_+$, with projection map $x\mapsto |x|$. Now $0\in \mathbb R_+$ has no neighbourhood that lifts to a homeomorphic neighbourhood in $\mathbb R$ ( a neighbourhood $[0,\epsilon)$ lifts to $(-\epsilon, \epsilon)$, which can't be written $\bigsqcup_{i\in I}U_i$, with $U_i\to [0, \epsilon)$ a homeomorphism)

So you can always form $E/G$, but the map $E\to E/G$ won't always be a covering map.

Now if you start from a connected covering $p:E\to Y$ then automatically $Aut(E,p)$ has no fixed points, by uniqueness of lifting, so that's not even a question; $E/Aut(E,p)$ might be different from $Y$ for other reasons though (if $p$ is not a regular covering)