In this proof we use the notation $x/U$ to denote an object $x$ of $\mathcal{S}$ which lies over $U$ in $\mathcal{C}$. Similarly $y/V$ denotes an object $y$ of $\mathcal{T}$ which lies over $V$ in $\mathcal{D}$. In the same vein $\alpha/a : x/U \to x'/U'$ denotes the morphism $\alpha : x \to x'$ with image $a : U \to U'$ in $\mathcal{C}$. \medskip\noindent Let $G : u_p\mathcal{S} \to \mathcal{T}$ be a $1$-morphism of fibred categories over $\mathcal{D}$. Denote $G' : u_{pp}\mathcal{S} \to \mathcal{T}$ the composition of $G$ with the canonical (localization) functor $u_{pp}\mathcal{S} \to u_p\mathcal{S}$. Then consider the functor $H : \mathcal{S} \to u^p\mathcal{T}$ given by $$ H(x/U) = (U, G'(U, \text{id}_{u(U)} : u(U) \to u(U), x)) $$ on objects and by $$ H((\alpha, a) : x/U \to x'/U') = G'(a, u(a), \alpha) $$ on morphisms. Since $G$ transforms strongly cartesian morphisms into strongly cartesian morphisms, we see that if $\alpha$ is strongly cartesian, then $H(\alpha)$ is strongly cartesian. Namely, we've seen in the proof of Lemma \ref{lemma-fibred-category-pullback} that in this case the map $(a, u(a), \alpha)$ becomes strongly cartesian in $u_p\mathcal{S}$. Clearly this construction is functorial in $G$ and we obtain a functor $$ A : \Mor_{\textit{Fib}/\mathcal{D}}(u_p\mathcal{S}, \mathcal{T}) \longrightarrow \Mor_{\textit{Fib}/\mathcal{C}}(\mathcal{S}, u^p\mathcal{T}) $$ \medskip\noindent Conversely, let $H : \mathcal{S} \to u^p\mathcal{T}$ be a $1$-morphism of fibred categories. Recall that an object of $u^p\mathcal{T}$ is a pair $(U, y)$ with $y \in \Ob(\mathcal{T}_{u(U)})$. We denote $\text{pr} : u^p\mathcal{T} \to \mathcal{T}$ the functor $(U, y) \mapsto y$. In this case we define a functor $G' : u_{pp}\mathcal{S} \to \mathcal{T}$ by the rules $$ G'(U, \phi : V \to u(U), x) = \phi^*\text{pr}(H(x)) $$ on objects and we let $$ G'((a, b, \alpha) : (U, \phi : V \to u(U), x) \to (U', \phi' : V' \to u(U'), x')) = \beta $$ be the unique morphism $\beta : \phi^*\text{pr}(H(x)) \to (\phi')^*\text{pr}(H(x'))$ such that $q(\beta) = b$ and the diagram $$ \xymatrix{ \phi^*\text{pr}(H(x)) \ar[d] \ar[r]_-{\beta} & (\phi')^*\text{pr}(H(x')) \ar[d] \\ \text{pr}(H(x)) \ar[r]^{\text{pr}(H(a, \alpha))} & \text{pr}(H(x')) } $$ Such a morphism exists and is unique because $\mathcal{T}$ is a fibred category. \medskip\noindent We check that $G'(r)$ is an isomorphism if $r \in R$. Namely, if $$ (a, \text{id}_V, \alpha) : (U', \phi' : V \to u(U'), x') \longrightarrow (U, \phi : V \to u(U), x) $$ with $\alpha$ strongly cartesian is an element of the right multiplicative system $R$ of Lemma \ref{lemma-right-multiplicative-system} then $H(\alpha)$ is strongly cartesian, and $\text{pr}(H(\alpha))$ is strongly cartesian, see proof of Lemma \ref{lemma-fibred-category-pushforward}. Hence in this case the morphism $\beta$ has $q(\beta) = \text{id}_V$ and is strongly cartesian. Hence $\beta$ is an isomorphism by Categories, Lemma \ref{categories-lemma-composition-cartesian}. Thus by Categories, Lemma \ref{categories-lemma-properties-right-localization} we obtain a canonical extension $G : u_p\mathcal{S} \to \mathcal{T}$. \medskip\noindent Next, let us prove that $G$ transforms strongly cartesian morphisms into strongly cartesian morphisms. Suppose that $f : X \to Y$ is a strongly cartesian. By the characterization of strongly cartesian morphisms in $u_p\mathcal{S}$ we can write $f$ as $((a, b, \alpha) : X' \to Y, r : X' \to Y)$ where $r \in R$ and $\alpha$ strongly cartesian in $\mathcal{S}$. By the above it suffices to show that $G'(a, b \alpha)$ is strongly cartesian. As before the condition that $\alpha$ is strongly cartesian implies that $\text{pr}(H(a, \alpha)) : \text{pr}(H(x)) \to \text{pr}(H(x'))$ is strongly cartesian in $\mathcal{T}$. Since in the commutative square above now all arrows except possibly $\beta$ is strongly cartesian it follows that also $\beta$ is strongly cartesian as desired. Clearly the construction $H \mapsto G$ is functorial in $H$ and we obtain a functor $$ B : \Mor_{\textit{Fib}/\mathcal{C}}(\mathcal{S}, u^p\mathcal{T}) \longrightarrow \Mor_{\textit{Fib}/\mathcal{D}}(u_p\mathcal{S}, \mathcal{T}) $$ To finish the proof of the lemma we have to show that the functors $A$ and $B$ are mutually quasi-inverse. We omit the verifications.