In Criteria for Representability, Lemma \ref{criteria-lemma-representable-by-spaces-limit-preserving} we have seen that (2) and (3) are equivalent. Thus it suffices to show that (1) and (2) are equivalent. One direction we saw in Lemma \ref{lemma-limit-preserving-objects}. For the other direction, let $U = \lim_{i \in I} U_i$ be the directed limit of affine schemes $U_i$ over $S$. We have to show that $$ \colim \mathcal{X}_{U_i} \longrightarrow \mathcal{X}_U \times_{\mathcal{Y}_U} \colim \mathcal{Y}_{U_i} $$ is an equivalence. Since we are assuming (2) we know that it is essentially surjective. Hence we need to prove it is fully faithful. Since $p$ is faithful on fibre categories (Algebraic Stacks, Lemma \ref{algebraic-lemma-criterion-map-representable-spaces-fibred-in-groupoids}) we see that the functor is faithful. Let $x_i$ and $x'_i$ be objects in the fibre category of $\mathcal{X}$ over $U_i$. The functor above sends $x_i$ to $(x_i|_U, p(x_i), can)$ where $can$ is the canonical isomorphism $p(x_i|_U) \to p(x_i)|_U$. Thus we assume given a morphism $$ (\alpha, \beta_i) : (x_i|_U, p(x_i), can) \longrightarrow (x'_i|_U, p(x'_i), can) $$ in the category of the right hand side of the first displayed arrow of this proof. Our task is to produce an $i' \geq i$ and a morphism $x_i|_{U_{i'}} \to x'_i|_{U_{i'}}$ which maps to $(\alpha, \beta_i|_{U_{i'}})$. \medskip\noindent Set $y_i = p(x_i)$ and $y'_i = p(x'_i)$. By (Algebraic Stacks, Lemma \ref{algebraic-lemma-criterion-map-representable-spaces-fibred-in-groupoids}) the functor $$ X_{y_i} : (\Sch/U_i)^{opp} \to \textit{Sets},\quad V/U_i \mapsto \{(x, \phi) \mid x \in \Ob(\mathcal{X}_V), \phi : f(x) \to y_i|V\}/\cong $$ is an algebraic space over $U_i$ and the same is true for the analogously defined functor $X_{y'_i}$. Since (2) is equivalent to (3) we see that $X_{y'_i}$ is locally of finite presentation over $U_i$. Observe that $(x_i, \text{id})$ and $(x'_i, \text{id})$ define $U_i$-valued points of $X_{y_i}$ and $X_{y'_i}$. There is a transformation of functors $$ \beta_i : X_{y_i} \to X_{y'_i},\quad (x/V, \phi) \mapsto (x/V, \beta_i|_V \circ \phi) $$ in other words, this is a morphism of algebraic spaces over $U_i$. We claim that $$ \xymatrix{ U \ar[d] \ar[rr] & & U_i \ar[d]^{(x'_i, \text{id})} \\ U_i \ar[r]^{(x_i, \text{id})} & X_{y_i} \ar[r]^{\beta_i} & X_{y'_i} } $$ commutes. Namely, this is equivalent to the condition that the pairs $(x_i|_U, \beta_i|_U)$ and $(x'_i|_U, \text{id})$ as in the definition of the functor $X_{y'_i}$ are isomorphic. And the morphism $\alpha : x_i|_U \to x'_i|_U$ exactly produces such an isomorphism. Arguing backwards the reader sees that if we can find an $i' \geq i$ such that the diagram $$ \xymatrix{ U_{i'} \ar[d] \ar[rr] & & U_i \ar[d]^{(x'_i, \text{id})} \\ U_i \ar[r]^{(x_i, \text{id})} & X_{y_i} \ar[r]^{\beta_i} & X_{y'_i} } $$ commutes, then we obtain an isomorphism $x_i|_{U_{i'}} \to x'_i|_{U_{i'}}$ which is a solution to the problem posed in the preceding paragraph. However, the diagonal morphism $$ \Delta : X_{y'_i} \to X_{y'_i} \times_{U_i} X_{y'_i} $$ is locally of finite presentation (Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-diagonal-morphism-finite-type}) hence the fact that $U \to U_i$ equalizes the two morphisms to $X_{y'_i}$, means that for some $i' \geq i$ the morphism $U_{i'} \to U_i$ equalizes the two morphisms, see Limits of Spaces, Proposition \ref{spaces-limits-proposition-characterize-locally-finite-presentation}.