[proofplan] We use the evenly covered neighbourhoods of the covering map as fibre-bundle trivializing neighbourhoods. On each such neighbourhood, an ordering of the sheets identifies $p^{-1}(U)$ with $U \times \{1,\dots,n\}$. On overlaps, two such orderings identify each sheet with exactly one sheet locally, so the transition map is given by a locally constant permutation-valued function. Since the fibre is discrete and $S_n$ acts on it by diffeomorphisms, these transition maps are smooth bundle transitions with structure group $S_n$. [/proofplan]

[step:Choose ordered sheets over each evenly covered neighbourhood] Let $F := \{1,\dots,n\}$, equipped with the discrete topology and its induced $0$-dimensional smooth manifold structure. Let $S_n$ denote the symmetric group of bijections $F \to F$, acting on $F$ by the evaluation map \begin{align*} S_n \times F &\to F, \quad (\sigma,i) \mapsto \sigma(i). \end{align*} Because $p$ is an $n$-sheeted covering map, every point of $B$ has exactly $n$ preimages and is contained in an evenly covered open neighbourhood. For each $b \in B$, choose an evenly covered open neighbourhood $U_a \subset B$ of $b$. Thus there exist open subsets $V_{a,1},\dots,V_{a,n} \subset E$ such that \begin{align*} p^{-1}(U_a) = \bigsqcup_{i=1}^{n} V_{a,i}, \end{align*} and each restricted map \begin{align*} p|_{V_{a,i}}: V_{a,i} \to U_a \end{align*} is a diffeomorphism. The index $a$ ranges over the chosen evenly covered neighbourhoods, and the index $i \in F$ records the chosen ordering of the sheets above $U_a$. [/step]

[step:Build local trivializations from the ordered sheets]For each index $a$, define a map \begin{align*} \Phi_a: p^{-1}(U_a) &\to U_a \times F \end{align*} by the rule \begin{align*} \Phi_a(e) := (p(e), i) \quad \text{whenever } e \in V_{a,i}. \end{align*} This is well-defined because the subsets $V_{a,1},\dots,V_{a,n}$ are pairwise disjoint and cover $p^{-1}(U_a)$. The inverse map is the map \begin{align*} \Psi_a: U_a \times F &\to p^{-1}(U_a), \quad (x,i) \mapsto (p|_{V_{a,i}})^{-1}(x). \end{align*} Indeed, if $e \in V_{a,i}$, then \begin{align*} \Psi_a(\Phi_a(e)) = \Psi_a(p(e),i) = (p|_{V_{a,i}})^{-1}(p(e)) = e, \end{align*} and if $(x,i) \in U_a \times F$, then \begin{align*} \Phi_a(\Psi_a(x,i)) = \Phi_a((p|_{V_{a,i}})^{-1}(x)) = (x,i). \end{align*} Since $F$ is discrete, $U_a \times F$ is the disjoint union of the open-and-closed components $U_a \times \{i\}$. On each component $U_a \times \{i\}$, the map $\Psi_a$ is exactly the diffeomorphism $(p|_{V_{a,i}})^{-1}: U_a \to V_{a,i}$, with the discrete factor recorded. Hence $\Psi_a$ is smooth, and its inverse $\Phi_a$ is smooth. Therefore $\Phi_a$ is a diffeomorphism. Finally, \begin{align*} \operatorname{pr}_1(\Phi_a(e)) = \operatorname{pr}_1(p(e),i) = p(e), \end{align*} where $\operatorname{pr}_1: U_a \times F \to U_a$ is the projection onto the first factor. Thus each $\Phi_a$ is a local trivialization of $p$ over $U_a$.[/step]

[guided]The goal of this step is to turn the covering-space data into the exact shape required for a fibre-bundle chart. A fibre-bundle trivialization over $U_a$ must be a diffeomorphism from $p^{-1}(U_a)$ to $U_a \times F$ whose first coordinate is $p$. Because $U_a$ is evenly covered, the preimage $p^{-1}(U_a)$ is split into ordered sheets: \begin{align*} p^{-1}(U_a) = \bigsqcup_{i=1}^{n} V_{a,i}, \end{align*} and each restriction $p|_{V_{a,i}}: V_{a,i} \to U_a$ is a diffeomorphism. This means that every point $e \in p^{-1}(U_a)$ lies in exactly one sheet $V_{a,i}$. We therefore define \begin{align*} \Phi_a: p^{-1}(U_a) &\to U_a \times F \end{align*} by \begin{align*} \Phi_a(e) := (p(e), i) \quad \text{if } e \in V_{a,i}. \end{align*} The disjointness of the sheets is what makes this definition unambiguous. To prove that $\Phi_a$ is a diffeomorphism, we write down its inverse explicitly as the map \begin{align*} \Psi_a: U_a \times F &\to p^{-1}(U_a), \quad (x,i) \mapsto (p|_{V_{a,i}})^{-1}(x). \end{align*} This is well-defined because $p|_{V_{a,i}}: V_{a,i} \to U_a$ is a diffeomorphism, hence has a smooth inverse $U_a \to V_{a,i}$. Now check the two compositions. If $e \in V_{a,i}$, then \begin{align*} \Psi_a(\Phi_a(e)) = \Psi_a(p(e),i) = (p|_{V_{a,i}})^{-1}(p(e)) = e. \end{align*} Conversely, if $(x,i) \in U_a \times F$, then $\Psi_a(x,i) \in V_{a,i}$ and $p(\Psi_a(x,i))=x$, so \begin{align*} \Phi_a(\Psi_a(x,i)) = (x,i). \end{align*} Thus $\Phi_a$ and $\Psi_a$ are inverse bijections. It remains to justify smoothness. Since $F$ is discrete, the product $U_a \times F$ is a disjoint union of the components $U_a \times \{i\}$. On the component $U_a \times \{i\}$, the inverse map $\Psi_a$ is exactly the smooth map $(p|_{V_{a,i}})^{-1}: U_a \to V_{a,i}$. Similarly, on the open subset $V_{a,i} \subset p^{-1}(U_a)$, the map $\Phi_a$ is the smooth map $e \mapsto (p(e),i)$. Hence both maps are smooth on each component of a disjoint open decomposition, so $\Phi_a$ is a diffeomorphism. Finally, the first coordinate of $\Phi_a(e)$ is $p(e)$: \begin{align*} \operatorname{pr}_1(\Phi_a(e)) = p(e). \end{align*} Therefore $\Phi_a$ is a valid fibre-bundle trivialization over $U_a$.[/guided]

[step:Compute the transition maps on overlaps as permutations] Let $U_a$ and $U_c$ be two chosen evenly covered open sets, and let \begin{align*} W_{ac} := U_a \cap U_c. \end{align*} Assume $W_{ac} \neq \varnothing$. The transition map over the overlap is \begin{align*} \Phi_c \circ \Phi_a^{-1}: W_{ac} \times F \to W_{ac} \times F. \end{align*} Fix $i \in F$. For each $x \in W_{ac}$, the point $\Phi_a^{-1}(x,i)$ lies in the fibre $p^{-1}(\{x\})$. Since the sets $V_{c,1},\dots,V_{c,n}$ partition $p^{-1}(U_c)$, there is a unique index $j \in F$ such that \begin{align*} \Phi_a^{-1}(x,i) \in V_{c,j}. \end{align*} Define \begin{align*} g_{ca}: W_{ac} &\to S_n \end{align*} as follows: for each $x \in W_{ac}$, the permutation $g_{ca}(x) \in S_n$ is determined by \begin{align*} g_{ca}(x)(i) = j \end{align*} exactly when $\Phi_a^{-1}(x,i) \in V_{c,j}$. For fixed $x \in W_{ac}$, the assignment $i \mapsto g_{ca}(x)(i)$ is a bijection $F \to F$, because both orderings list the same $n$ points of the fibre $p^{-1}(\{x\})$ exactly once. Therefore $g_{ca}(x) \in S_n$. By construction, \begin{align*} (\Phi_c \circ \Phi_a^{-1})(x,i) = (x, g_{ca}(x)(i)) \end{align*} for all $(x,i) \in W_{ac} \times F$. [/step]

[step:Show the permutation-valued transition functions are locally constant] We prove that each map $g_{ca}: W_{ac} \to S_n$ is locally constant. Fix $x_0 \in W_{ac}$ and $i \in F$, and set \begin{align*} j_0 := g_{ca}(x_0)(i). \end{align*} By definition, the point $\Phi_a^{-1}(x_0,i)$ belongs to the [open set](/page/Open%20Set) $V_{a,i} \cap V_{c,j_0}$. Since $V_{a,i} \cap V_{c,j_0}$ is open in $E$, the subset $V_{a,i} \cap V_{c,j_0}$ is open in the subspace $V_{a,i}$. Since $p|_{V_{a,i}}: V_{a,i} \to U_a$ is a diffeomorphism, hence an open map, the image \begin{align*} N := p(V_{a,i} \cap V_{c,j_0}) \end{align*} is open in $U_a$. Moreover $V_{c,j_0} \subset p^{-1}(U_c)$, so $N \subset U_c$; therefore $N$ is open in the overlap $W_{ac}=U_a \cap U_c$. It contains $x_0$ because $\Phi_a^{-1}(x_0,i) \in V_{a,i} \cap V_{c,j_0}$. For every $x \in N$, the point $\Phi_a^{-1}(x,i)$ lies in $V_{c,j_0}$, so \begin{align*} g_{ca}(x)(i) = j_0. \end{align*} Because $F$ is finite, intersecting the corresponding neighbourhoods for all $i \in F$ gives an open neighbourhood $N_0 \subset W_{ac}$ of $x_0$ on which $g_{ca}(x)(i)=g_{ca}(x_0)(i)$ for every $i \in F$. Hence \begin{align*} g_{ca}(x) = g_{ca}(x_0) \end{align*} for all $x \in N_0$. Thus $g_{ca}: W_{ac} \to S_n$ is locally constant. [/step]

[step:Conclude that the covering is a smooth fibre bundle with structure group $S_n$] The maps $\Phi_a: p^{-1}(U_a) \to U_a \times F$ form local trivializations of $p$ over the open cover $\{U_a\}$ of $B$. On each nonempty overlap $W_{ac}=U_a \cap U_c$, the corresponding transition map has the form \begin{align*} (\Phi_c \circ \Phi_a^{-1})(x,i) = (x, g_{ca}(x)(i)), \end{align*} where $g_{ca}: W_{ac} \to S_n$ is locally constant. Since $S_n$ is discrete and $F$ is a discrete smooth manifold, every locally constant map into $S_n$ is smooth, and every element of $S_n$ acts on $F$ by a diffeomorphism. Therefore the transition maps are smooth fibre-bundle transition maps with values in the structure group $S_n$. Hence $p: E \to B$ is a smooth fibre bundle with fibre $F=\{1,\dots,n\}$ and structure group $S_n$. [/step]