[proofplan] We first show that the prescribed topology makes each domain $\pi^{-1}(U_i)$ open in $E$ and makes each $\Phi_i$ a homeomorphism onto $U_i\times F$. We then use the maps $\Phi_i$, together with ordinary product charts on $U_i\times F$, to define a smooth atlas on $E$; the assumed smoothness of the transition maps is exactly the compatibility condition for this atlas. In these charts, $\pi$ is the projection $U_i\times F\to U_i$, so it is smooth and the maps $\Phi_i$ are smooth local trivializations. Finally, uniqueness follows because any smooth structure making all $\Phi_i$ diffeomorphisms must contain exactly the same pulled-back product smooth structure. [/proofplan]

[step:Show that the local trivialization domains are open and that the trivializations are homeomorphisms]For each $i\in I$, define \begin{align*} E_i:=\pi^{-1}(U_i)\subset E. \end{align*} We first prove that $E_i$ is open in the prescribed topology. For every $j\in I$, \begin{align*} \Phi_j(E_i\cap E_j) = (U_i\cap U_j)\times F, \end{align*} because each $\Phi_j$ preserves the projection to $B$. Since $U_i\cap U_j$ is open in $B$, the set $(U_i\cap U_j)\times F$ is open in $U_j\times F$. Hence $E_i$ is open by the definition of the topology on $E$. We next prove that \begin{align*} \Phi_i:E_i&\to U_i\times F \end{align*} is a homeomorphism. It is bijective by hypothesis. If $W\subset E_i$ is open in the [subspace topology](/page/Subspace%20Topology), then $W=E_i\cap W_0$ for some [open set](/page/Open%20Set) $W_0\subset E$, and therefore $\Phi_i(W)$ is open in $U_i\times F$ by the definition of the topology. Conversely, let $A\subset U_i\times F$ be open and define \begin{align*} W:=\Phi_i^{-1}(A)\subset E_i. \end{align*} For every $j\in I$, the transition map \begin{align*} T_{ij}:=\Phi_i\circ \Phi_j^{-1}:(U_i\cap U_j)\times F&\to (U_i\cap U_j)\times F \end{align*} is a smooth diffeomorphism by hypothesis, hence in particular a homeomorphism. Therefore \begin{align*} \Phi_j(W\cap E_j) &= T_{ij}^{-1}\left(A\cap ((U_i\cap U_j)\times F)\right), \end{align*} which is open in $U_j\times F$. Thus $W$ is open in $E$. This proves that $\Phi_i$ is a homeomorphism.[/step]

[guided]The first point is topological: before constructing a smooth structure, we must know that each proposed chart domain is open. For each $i\in I$, set \begin{align*} E_i:=\pi^{-1}(U_i). \end{align*} To test whether $E_i$ is open in the topology on $E$, we must inspect its image under every local trivialization $\Phi_j$. Since $\Phi_j$ preserves the projection, an element $e\in E_j$ lies in $E_i$ exactly when $\pi(e)\in U_i$. Under $\Phi_j$, this says precisely that the first coordinate belongs to $U_i\cap U_j$. Hence \begin{align*} \Phi_j(E_i\cap E_j)=(U_i\cap U_j)\times F. \end{align*} Because $U_i$ and $U_j$ are open in $B$, their intersection is open in $B$, and therefore $(U_i\cap U_j)\times F$ is open in the product manifold $U_j\times F$. By the defining rule for the topology on $E$, this proves that $E_i$ is open. Now we verify that $\Phi_i:E_i\to U_i\times F$ is a homeomorphism. It is already a bijection by hypothesis, so only continuity and openness need to be checked. If $W\subset E_i$ is open in the subspace topology, then $W=E_i\cap W_0$ for some open set $W_0\subset E$. Applying the definition of the topology to $W_0$ with the index $i$ gives that $\Phi_i(W)$ is open in $U_i\times F$. Thus $\Phi_i$ is an open map. For continuity, take an open set $A\subset U_i\times F$ and define \begin{align*} W:=\Phi_i^{-1}(A)\subset E_i. \end{align*} We must prove that $W$ is open in $E$. The topology on $E$ tells us to check all trivializations $\Phi_j$. On the overlap $E_i\cap E_j$, define the transition map \begin{align*} T_{ij}:=\Phi_i\circ \Phi_j^{-1}:(U_i\cap U_j)\times F\to (U_i\cap U_j)\times F. \end{align*} This map is a smooth diffeomorphism by hypothesis, so it is a homeomorphism. Therefore \begin{align*} \Phi_j(W\cap E_j) = T_{ij}^{-1}\left(A\cap ((U_i\cap U_j)\times F)\right). \end{align*} The set $A\cap ((U_i\cap U_j)\times F)$ is open in $(U_i\cap U_j)\times F$, and its inverse image under the homeomorphism $T_{ij}$ is open. Hence $\Phi_j(W\cap E_j)$ is open for every $j\in I$, so $W$ is open in $E$. This proves that $\Phi_i$ is continuous. Since $\Phi_i$ is bijective, continuous, and open, it is a homeomorphism.[/guided]

[step:Construct the smooth atlas by pulling back product charts] Let $m\in\mathbb{N}$ denote the dimension of the smooth manifold $B$, and let $n\in\mathbb{N}$ denote the dimension of the smooth manifold $F$. Let $(V,\alpha)$ be a smooth chart on $B$ with $V\subset U_i$, where \begin{align*} \alpha:V\to \alpha(V)\subset \mathbb{R}^m \end{align*} is a diffeomorphism onto an open subset of $\mathbb{R}^m$. Such charts cover $U_i$ because any smooth chart $(\widetilde V,\widetilde\alpha)$ on $B$ may be restricted to the open set $V:=\widetilde V\cap U_i$, and $\widetilde\alpha|_V$ is again a smooth chart. Let $(Z,\beta)$ be a smooth chart on $F$, where \begin{align*} \beta:Z\to \beta(Z)\subset \mathbb{R}^n \end{align*} is a diffeomorphism onto an open subset of $\mathbb{R}^n$. Define the corresponding chart on $E$ as the map \begin{align*} \theta_{i,V,Z}:\Phi_i^{-1}(V\times Z)\to \alpha(V)\times \beta(Z)\subset \mathbb{R}^{m+n} \end{align*} given by \begin{align*} \theta_{i,V,Z}(e)=(\alpha\times \beta)(\Phi_i(e)) \end{align*} for every $e\in\Phi_i^{-1}(V\times Z)$. The domain $\Phi_i^{-1}(V\times Z)$ is open in $E$ because $\Phi_i$ is a homeomorphism and $V\times Z$ is open in $U_i\times F$. Since the sets $U_i$ cover $B$, the restricted charts on $B$ cover each $U_i$, and the charts on $F$ cover $F$, these charts cover $E$. Let $\theta_{i,V,Z}$ and $\theta_{j,V',Z'}$ be two such charts with overlapping domains. Here $(V',\alpha')$ is a smooth chart on $B$ with $V'\subset U_j$ and \begin{align*} \alpha':V'\to \alpha'(V')\subset \mathbb{R}^m \end{align*} is a diffeomorphism onto an open subset of $\mathbb{R}^m$, while $(Z',\beta')$ is a smooth chart on $F$ and \begin{align*} \beta':Z'\to \beta'(Z')\subset \mathbb{R}^n \end{align*} is a diffeomorphism onto an open subset of $\mathbb{R}^n$. On their overlap, the coordinate change is \begin{align*} \theta_{i,V,Z}\circ \theta_{j,V',Z'}^{-1} = (\alpha\times \beta)\circ \Phi_i\circ \Phi_j^{-1}\circ(\alpha'\times \beta')^{-1}, \end{align*} restricted to the appropriate open subset of $\alpha'(V')\times\beta'(Z')$. The maps $\alpha,\alpha',\beta,\beta'$ are smooth coordinate maps, and $\Phi_i\circ\Phi_j^{-1}$ is smooth by hypothesis. Hence every coordinate change is smooth. The inverse coordinate change is smooth by the same argument, or equivalently because $\Phi_i\circ\Phi_j^{-1}$ is a diffeomorphism. Thus the charts $\theta_{i,V,Z}$ form a smooth atlas on $E$. Because the topology on $E$ is Hausdorff and second-countable by hypothesis, this atlas defines a smooth manifold structure on $E$. [/step]

[step:Verify that the trivializations are diffeomorphisms and that $\pi$ is a smooth fibre bundle projection] By construction, each map \begin{align*} \Phi_i:E_i&\to U_i\times F \end{align*} is a homeomorphism, and the smooth structure on $E_i$ is the pullback of the product smooth structure on $U_i\times F$ through $\Phi_i$. Therefore $\Phi_i$ is a diffeomorphism. For $e\in E_i$, the projection-preserving hypothesis gives \begin{align*} \pi(e)=\operatorname{pr}_1(\Phi_i(e)). \end{align*} Thus, on $E_i$, \begin{align*} \pi|_{E_i}=\operatorname{pr}_1\circ \Phi_i. \end{align*} The map $\Phi_i$ is smooth, and the projection \begin{align*} \operatorname{pr}_1:U_i\times F&\to U_i \end{align*} is smooth for the product smooth structure. Hence $\pi|_{E_i}$ is smooth for every $i\in I$. Since the open sets $E_i$ cover $E$, the map $\pi:E\to B$ is smooth. Finally, each $\Phi_i$ is a diffeomorphism satisfying $\operatorname{pr}_1\circ \Phi_i=\pi|_{E_i}$. Therefore $(E,\pi,B,F)$ is a smooth fibre bundle with local trivializations $(\Phi_i)_{i\in I}$. [/step]

[step:Prove uniqueness of the smooth manifold structure] Suppose $\mathcal{S}$ is another smooth manifold structure on the same [topological space](/page/Topological%20Space) $E$ such that every map \begin{align*} \Phi_i:E_i&\to U_i\times F \end{align*} is a diffeomorphism. Then every chart of the form \begin{align*} (\alpha\times\beta)\circ\Phi_i:\Phi_i^{-1}(V\times Z)&\to \alpha(V)\times\beta(Z) \end{align*} constructed above is a smooth chart for $\mathcal{S}$, because it is the composition of the diffeomorphism $\Phi_i$ with a product chart on $U_i\times F$. Conversely, since the sets $E_i$ cover $E$ and each $\Phi_i$ is a diffeomorphism for $\mathcal{S}$, any $\mathcal{S}$-smooth chart is smoothly compatible locally with the pulled-back product charts above. Hence the maximal smooth atlas determined by $\mathcal{S}$ is exactly the maximal atlas generated by the charts $\theta_{i,V,Z}$. Therefore $\mathcal{S}$ equals the smooth structure constructed above. This proves both existence and uniqueness of the smooth manifold structure and completes the proof. [/step]