I'm currently struggling with understanding the connection between Galois groups and fundamental groups for which I started reading a book called Galois Groups and Fundamental Groups by Tamás Szamuely. Now I'm stuck in something important:
Szamuely says that given a cover $p:Y\to X $, the group $\operatorname{Aut}(Y|X) $ acts naturally on $ Y$ i.e. if $ y \in Y$ and $ \varphi \in \operatorname{Aut}(Y|X) $, $ \varphi \cdot y= \varphi(y)$, and that the orbits of this action are the fibres of $ p$ (I understand that). Then he defines a function $\overline{p}:\operatorname{Aut}(Y|X)/Y \to X $ given by assigning each equivalence class $[x] $ the element $p(x) $, which is well defined since the equivalence classes are the orbits. The thing which I can't figure out is the definition of a Galois covering since Szamuely says that:
$Y$ is a Galois covering if $ \overline{p}$ is a homeomorphism, which is senseless to my because we cannot talk about continuity without assigning a topology to $ \operatorname{Aut}(Y|X)$.
This book is pretty advanced so I shall think that I'm missing something of the prerequisites. So my question is which is the "standard" topology given to $ \operatorname{Aut}(Y|X)$? I can assure that there is nothing about this information previously in the book.
Note: I can't provide a link to the book for I can't find it (I have it physically). If someone knows a link, it would be great if she/he paste it in the comments.
Thanks in advance
It is not necessary to put a topology on $\operatorname{Aut}(Y|X)$. Note that the space being considered is not $\operatorname{Aut}(Y|X)/Y$ (which would not make sense) but rather $\operatorname{Aut}(Y|X)\backslash Y$, the set of (left) orbits of $\operatorname{Aut}(Y|X)$ on $Y$. That is, it is the quotient of $Y$ by the equivalence relation whose equivalence classes are orbits of the action. No topology on $\operatorname{Aut}(Y|X)$ is needed to put a topology on this quotient, just a topology on $Y$ (so you can then use the quotient topology).