**Proof.**
In Criteria for Representability, Lemma 95.5.3 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 100.3.2. For the other direction, let $U = \mathop{\mathrm{lim}}\nolimits _{i \in I} U_ i$ be the directed limit of affine schemes $U_ i$ over $S$. We have to show that

\[ \mathop{\mathrm{colim}}\nolimits \mathcal{X}_{U_ i} \longrightarrow \mathcal{X}_ U \times _{\mathcal{Y}_ U} \mathop{\mathrm{colim}}\nolimits \mathcal{Y}_{U_ i} \]

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 92.9.2) 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 $(\alpha , \beta _ i|_{U_{i'}})$.

Set $y_ i = p(x_ i)$ and $y'_ i = p(x'_ i)$. By (Algebraic Stacks, Lemma 92.9.2) the functor

\[ X_{y_ i} : (\mathit{Sch}/U_ i)^{opp} \to \textit{Sets},\quad V/U_ i \mapsto \{ (x, \phi ) \mid x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ V), \phi : f(x) \to y_ i|V\} /\cong \]

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 65.28.10) 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 68.3.10.
$\square$

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