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This is a question based on the heuristics that most things in algebraic/differential topology has an analogue in algebraic geometry.

The fundamental group classifies the covering spaces of a (pointed, connected, path connected, semilocally simply connected topological) space. First we construct the universal cover as a space of paths and equip it with compact-open topology. And then we use the action of the fundamental group on the cover to define the space as a quotient. And by actions of the subgroups, we construct all the covering spaces, which are in one-one correspondence with the conjugacy classes of subgroups of fundamental group of the pointed space.

The question is, up to what extent can we carry over this setup to algebraic geometry? There are the appropriate notions of etale coverings and etale fundamental groups already, and by GAGA for smooth algebraic varieties one can already see some hope.

So, are there algebraic geometric analogues of the theorem on existence of universal cover and the theorem classification of covering spaces in one-one correspondence with conjugacy classes of subgroups of the fundamental groups?

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    $\begingroup$ In para. 3 you meant "algebraic geometry"? Deep result of Grauert/Remmert: if $X$ is loc. f.type over $\mathbf{C}$ (no smoothness, no properness), analytification is equivalence from category of finite etale covers of $X$ to finite covering spaces of $X(\mathbf{C})$. Best analogue of universal cover: invlim $X'$ of pointed conn'd finite etale covers (for $X$ conn'd of f.type, say). This gigantic scheme is conn'd, no nontrivial conn'd finite etale covers, ${\rm{Aut}}(X'/X)$ is "opposite" to $\pi_1(X,x)$, & conj. classes of closed subgps correspond to conn'd "ind-finite-etale" covers of $X$. $\endgroup$
    – BCnrd
    Commented May 11, 2010 at 18:47
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    $\begingroup$ An instructive example beyond the "limit of finite etale" case: let $X$ be the nodal plane cubic. The infinite tree $X'$ of projective lines is etale over $X$ with covering group $\mathbf{Z}$, and for each $n \ge 1$ there's a unique $n$-gon intermediate cover $X_n$. This $X'$ is not "ind-finite-etale" over $X$, and is the analytification of the topological universal cover. So one has the "surprise" result that ${\rm{H}}^1(X,\mathbf{Z}) \ne 0$ (etale cohomology) whereas $\pi_1(X)$ has no nonzero cont. homs to $\mathbf{Z}$. Upshot: for non-normal $X$ some funny things happen. $\endgroup$
    – BCnrd
    Commented May 11, 2010 at 18:52
  • $\begingroup$ See also <a href="mathoverflow.net/questions/21717/… answer</a>, and Ravi Vakhil's comment. $\endgroup$
    – Charles Staats
    Commented May 11, 2010 at 20:29

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Yes, (and of course): The very definition of the étale fundamental group is that it classifies finite étale covers.

Precisely: Let $X$ be a connected scheme and $x$ be a geometric point of $X$. There is by construction an equivalence of categories between finite $\pi_1^{\mathrm{ét}}(X,x)$--sets and finite étale coverings of $X$, connected coverings corresponding to transitive sets. This property characterises $\pi_1^{\mathrm{ét}}(X,x)$ up to unique isomorphism.

A universal cover as in topology also exists, although only as a profinite cover.

A fine source for this is T. Szamuely's book "Fundamental groups and Galois groups" (Cambridge 2009).

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