A covering map $f:X\rightarrow Y$ is called Galois if for each $y\in Y$ and each pair of lifts $x, x^{'}$, there is a covering transformation taking $x$ to $x^{'}$. What is a good way to understand this definition? It seems to me that $f$ is Galois if and only if $Y$ is obtained from $X$ as a quotient of some group.
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7$\begingroup$ Indeed. In fact $Y$ will be the quotient of $X$ by the group of deck transformations of $f$. (This is exactly like the case of a Galois extension of fields.) $\endgroup$– Zhen LinNov 29, 2012 at 20:15
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In the setting of (complex) algebraic geometry, the covering is Galois if and only if the function field $K(X)$ is a Galois extension of the function field $K(Y)$. Moreover, if $f$ is Galois, then the Galois group of the extension is exactly the deck transformation group $G$. As you've already noticed. If $f$ is Galois, then $Y$ is isomorphic to $X/G$, where $G$ is the Galois group.
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$\begingroup$ yes there is a very nice book concerning this subject @Hodge-Tate: Galois Groups and Fundamental Groups by Szamuely $\endgroup$ Nov 11, 2018 at 11:02
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$\begingroup$ where in this reference (Szamuely) is this correspondence Galois extension and Galois covering stated? My assumptions are $X$ and $Y$ Are Noetherian, separated, integral schemes over $\mathbb C$ (or at least varieties with base field $\mathbb C$) $\endgroup$– quantumSep 7, 2020 at 7:03