Proof

With the standard definitions of the two functors, both are in fact inverse isomorphisms of categories instead of just equivalences of categories (meaning that the required natural isomorphisms from the composites of the two functors to the identity functor are componentwise equalities), which establishes the claim right away. For definiteness, we make this explicit:

Given $\rho \in Set^{\Pi_1(X)}$ a permutation representation, we need to exhibit a natural isomorphism of permutation representations.

First consider what the right hand side is like: By this def. of $Rec$ and this def. of $Fib$ we have for every $x \in X$ an actual equality

To similarly understand the value of $Fib(Rec(\rho))$ on morphisms $[\gamma] \in \Pi_1(X)$, let $\gamma \colon [0,1] \to X$ be a representing path in $X$. As in the proof of the path lifting lemma for covering spaces (this lemma) we find a finite number of paths $\{\gamma_i\}_{i \in \{1,n\}}$ such that

1. regarded as morphisms $[\gamma_i]$ in $\Pi_1(X)$ they compose to $[\gamma]$:

   $$[\gamma] = [\gamma_n] \circ \cdots \circ [\gamma_2] \circ [\gamma_1]$$

2. each $\gamma_i$ factors through an open subset $U_i \subset X$ over which $Rec(\rho)$ trivializes.

Hence by functoriality of $Fib(Rec(\rho))$ it is sufficient to understand its value on these paths $\gamma_i$. But on these we have again by direct unwinding of the definitions that

This means that if we take

to be the above identification, then this is a natural transformation and hence in a particular a natural isomorphism, as required.

It remains to see that these morphisms $\eta_\rho$ are themselves natural in $\rho$, hence that for each morphism $\phi \colon \rho \to \rho'$ the diagram

commutes as a diagram in $Rep(\Pi_1(X), Set)$. Since these morphisms are themselves groupoid homotopies (natural isomorphisms) this is the case precisely if for all $x \in X$ the corresponding component diagram commutes. But by the above this is

and hence this means that the top and bottom horizontal morphism are in fact equal. Direct unwinding of the definitions shows that this is indeed the case.

Conversely, given $E \in Cov(X)$ a covering space, we need to exhibit a natural isomorphism of covering spaces of the form

Again by this def. of $Rec$ and this def. of $Fib$ the underlying set of $Rec(Fib(E))$ is actually equal to that of $E$, hence it is sufficient to check that this identity function on underlying sets is a homeomorphism of topological spaces.

By the assumption that $X$ is locally path-connected and semi-locally simply connected, it is sufficient to check for $U\subset X$ an open path-connected subset and $x \in X$ a point with the property that $\pi_1(U,x) \to \pi_1(X,x)$ lands is constant on the trivial element, that the open subsets of $E$ of the form $U \times \{\hat x\} \subset p^{-1}(U)$ form a basis for the topology of $Rec(Fib(E))$. But this is the case by definition of $Rec$.

It remains to see that $\epsilon_E$ is itself natural in $E$. But as for the converse direction, since the components of $\epsilon_E$ are in fact equalities, this follows by direct unwinding of the definitions.

This establishes an equivalence as required. In fact this is an adjoint equivalence.