This means we have to show the following: Given \begin{enumerate} \item an affine scheme $U = \lim_i U_i$ which is written as the directed limit of affine schemes $U_i$ over $S$, \item an object $y_i$ of $\mathcal{Y}$ over $U_i$ for some $i$, and \item an object $\Xi = (U, Z, y, x, \alpha)$ of $\mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ over $U$ such that $y = y_i|_U$, \end{enumerate} then there exists an $i' \geq i$ and an object $\Xi_{i'} = (U_{i'}, Z_{i'}, y_{i'}, x_{i'}, \alpha_{i'})$ of $\mathcal{H}_d(\mathcal{X}/\mathcal{Y})$ over $U_{i'}$ with $\Xi_{i'}|_U = \Xi$ and $y_{i'} = y_i|_{U_{i'}}$. Namely, the last two equalities will take care of the commutativity of (\ref{equation-limit-preserving}). \medskip\noindent Let $X_{y_i} \to U_i$ be an algebraic space representing the $2$-fibre product $$ (\Sch/U_i)_{fppf} \times_{y_i, \mathcal{Y}, F} \mathcal{X}. $$ Note that $X_{y_i} \to U_i$ is locally of finite presentation by our assumption on $F$. Write $\Xi $. It is clear that $\xi = (Z, Z \to U_i, x, \alpha)$ is an object of the $2$-fibre product displayed above, hence $\xi$ gives rise to a morphism $f_\xi : Z \to X_{y_i}$ of algebraic spaces over $U_i$ (since $X_{y_i}$ is the functor of isomorphisms classes of objects of $(\Sch/U_i)_{fppf} \times_{y, \mathcal{Y}, F} \mathcal{X}$, see Algebraic Stacks, Lemma \ref{algebraic-lemma-characterize-representable-by-space}). By Limits, Lemmas \ref{limits-lemma-descend-finite-presentation} and \ref{limits-lemma-descend-finite-locally-free} there exists an $i' \geq i$ and a finite locally free morphism $Z_{i'} \to U_{i'}$ of degree $d$ whose base change to $U$ is $Z$. By Limits of Spaces, Proposition \ref{spaces-limits-proposition-characterize-locally-finite-presentation} we may, after replacing $i'$ by a bigger index, assume there exists a morphism $f_{i'} : Z_{i'} \to X_{y_i}$ such that $$ \xymatrix{ Z \ar[d] \ar[r] \ar@/^3ex/[rr]^{f_\xi} & Z_{i'} \ar[d] \ar[r]_{f_{i'}} & X_{y_i} \ar[d] \\ U \ar[r] & U_{i'} \ar[r] & U_i } $$ is commutative. We set $\Xi_{i'} = (U_{i'}, Z_{i'}, y_{i'}, x_{i'}, \alpha_{i'})$ where \begin{enumerate} \item $y_{i'}$ is the object of $\mathcal{Y}$ over $U_{i'}$ which is the pullback of $y_i$ to $U_{i'}$, \item $x_{i'}$ is the object of $\mathcal{X}$ over $Z_{i'}$ corresponding via the $2$-Yoneda lemma to the $1$-morphism $$ (\Sch/Z_{i'})_{fppf} \to \mathcal{S}_{X_{y_i}} \to (\Sch/U_i)_{fppf} \times_{y_i, \mathcal{Y}, F} \mathcal{X} \to \mathcal{X} $$ where the middle arrow is the equivalence which defines $X_{y_i}$ (notation as in Algebraic Stacks, Sections \ref{algebraic-section-representable-by-algebraic-spaces} and \ref{algebraic-section-split}). \item $\alpha_{i'} : y_{i'}|_{Z_{i'}} \to F(x_{i'})$ is the isomorphism coming from the $2$-commutativity of the diagram $$ \xymatrix{ (\Sch/Z_{i'})_{fppf} \ar[r] \ar[rd] & (\Sch/U_i)_{fppf} \times_{y_i, \mathcal{Y}, F} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d]^F \\ & (\Sch/U_{i'})_{fppf} \ar[r] & \mathcal{Y} } $$ \end{enumerate} Recall that $f_\xi : Z \to X_{y_i}$ was the morphism corresponding to the object $\xi = (Z, Z \to U_i, x, \alpha)$ of $(\Sch/U_i)_{fppf} \times_{y_i, \mathcal{Y}, F} \mathcal{X}$ over $Z$. By construction $f_{i'}$ is the morphism corresponding to the object $\xi_{i'} = (Z_{i'}, Z_{i'} \to U_i, x_{i'}, \alpha_{i'})$. As $f_\xi = f_{i'} \circ (Z \to Z_{i'})$ we see that the object $\xi_{i'} = (Z_{i'}, Z_{i'} \to U_i, x_{i'}, \alpha_{i'})$ pulls back to $\xi$ over $Z$. Thus $x_{i'}$ pulls back to $x$ and $\alpha_{i'}$ pulls back to $\alpha$. This means that $\Xi_{i'}$ pulls back to $\Xi$ over $U$ and we win.