Finite étale covers are equivalent to finite continuous étale fundamental group sets

Choose a set of pointed representatives $(T_i,t_i)$ for the connected Galois covers. Write $i\ge j$ when there is a pointed map $u_{ij}:T_i\to T_j$; it is unique by [F2]. The relation is a directed partial order up to pointed isomorphism: antisymmetry follows because surjections in both directions give equal finite degrees, hence degree-one maps and isomorphisms; directedness follows by taking the component of $T_i{\times }_X{T}_j$ through $(t_i,t_j)$ and then a pointed Galois refinement of it. A Galois refinement of a disconnected cover trivializes all its components at once by the ordered-fibre construction in [F2]. Thus for every finite cover $Y$, evaluation yields a natural bijection ${<mrow><mi\; mathvariant="normal">c</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">l</mi><mi\; mathvariant="normal">i</mi><mi\; mathvariant="normal">m</mi></mrow>}_i{\mathrm{Hom}}_X(T_i,Y)⟶F(Y),a⟼a(t_i)\mathrm{.}$ It is onto by such a trivializing refinement. If two maps evaluate to the same point, take a common refinement and apply the connected-source uniqueness in [F2]; they are equal in the colimit.

Put $H_i={\mathrm{Aut}}_X(T_i)$. For $i\ge j$ and $h\in H_i$, simple transitivity gives a unique $h_j\in H_j$ with $h_j(t_j)=u_{ij}(h(t_i))$. Both maps $h_j{u}_{ij}$ and $u_{ij}h$ agree at $t_i$, hence agree everywhere by [F2]. This defines a homomorphism $H_i\to H_j$. It is surjective because $u_{ij}$ is surjective on the geometric fibre. The maps compose compatibly. Let and $G=H^{\mathrm{o}\mathrm{p}}$. The inverse limit is a closed subgroup of a product of finite discrete groups, hence compact, Hausdorff and totally disconnected by [F3]. Its projections onto $H_i$ are surjective: imposing any prescribed coordinate together with finitely many compatibility constraints can be solved at a common upper index, and compactness makes the resulting family of closed conditions simultaneously satisfiable.

A compatible family $h=(h_i)$ acts on the colimit in step 1.1 by precomposition $a\mapsto a{h}_i$. Precomposition reverses multiplication, so this gives a homomorphism $G\to \mathrm{Aut}(F)$. Its restriction on $F(T_i)$, identified with $H_i$ by evaluation at $t_i$, is right multiplication by $h_i$. Conversely, any natural automorphism $\gamma$ determines $h_i\in H_i$ by $h_i(t_i)={\gamma }_{T_i}(t_i)$. Naturality for $u_{ij}$ makes these compatible. Naturality for every map $a:T_i\to Y$ forces ${\gamma }_Y(a(t_i))=a(h_i(t_i))$, which determines $\gamma$ on every fibre by step 1.1. Thus $G\cong \mathrm{Aut}(F)$. This is a topological isomorphism: each finite fibre is represented at a single trivializing refinement, so its action factors through $H_i^{\mathrm{o}\mathrm{p}}$; conversely the action on $F(T_i)$ detects the entire $i$th coordinate. Hence both topologies have the same finite-coordinate neighbourhood basis. In particular the group in [F1] is profinite.

The group acts transitively on the fibre of every nonempty connected cover. Indeed choose a pointed Galois refinement surjecting onto that cover; the projection $H\to H_i$ is onto by step 2.1, and its regular action on $F(T_i)$ is transitive, so the induced action on the target fibre is transitive. For a general cover its connected decomposition is carried to its orbit decomposition: the group preserves every component by naturality for its inclusion, and acts transitively within it.

A continuous action on a finite set $E$ has an open kernel: intersect its finitely many open point stabilizers. By the inverse-limit topology of step 3.1, that kernel contains the kernel of $G\to H_i^{\mathrm{o}\mathrm{p}}$ for some $i$; finitely many coordinate constraints can be combined at one upper index. The projection is onto by step 2.1, so the action factors through $H_i^{\mathrm{o}\mathrm{p}}$. The contracted-cover construction in [F2] gives a finite étale cover with precisely that action on its fibre. This proves essential surjectivity.

Let $q:F(Y)\to F(Z)$ be equivariant. Its graph is an invariant subset of $F(Y{\times }_{X}Z)$, and hence, by step 4.1, a union of fibres of connected components of $Y{\times }_{X}Z$. Take the corresponding open and closed union of components $W$. Its projection to $Y$ is bijective on the fibre and hence an isomorphism by [F2]. The composite $Y\cong W\to Z$ has fibre map $q$. Faithfulness follows from [F2]; therefore the functor is fully faithful.

Steps 4.2 and 5.1 prove the equivalence, step 3.1 proves profiniteness, and step 4.1 identifies the connected covers. AC enters in choosing the pointed set of representatives and through [F2], and its additional compactness use is exactly step 2.1 via [F3]. This proof supplies reconstruction explicitly and does not invoke an unproved Galois-category theorem or a universal cover as an actual finite scheme.