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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    $\begingroup$ 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 Lin
    Nov 29, 2012 at 20:15

1 Answer 1


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.

  • $\begingroup$ Any book references for this? $\endgroup$
    – Hodge-Tate
    Oct 1, 2018 at 18:07
  • $\begingroup$ yes there is a very nice book concerning this subject @Hodge-Tate: Galois Groups and Fundamental Groups by Szamuely $\endgroup$
    – Simonsays
    Nov 11, 2018 at 11:02
  • $\begingroup$ 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 $\mathbb C$) $\endgroup$
    – quantum
    Sep 7, 2020 at 7:03

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