It is clear that (3) implies (2). \medskip\noindent Let us prove that (2) implies (1). Assume (2). Choose any affine opens $U \subset X$ and $V \subset S$ such that $f(U) \subset V$. We have to show that $\mathcal{O}_S(V) \to \mathcal{O}_X(U)$ is of finite presentation. Let $(A_i, \varphi_{ii'})$ be a directed system of $\mathcal{O}_S(V)$-algebras. Set $A = \colim_i A_i$. According to Algebra, Lemma \ref{algebra-lemma-characterize-finite-presentation} we have to show that $$ \Hom_{\mathcal{O}_S(V)}(\mathcal{O}_X(U), A) = \colim_i \Hom_{\mathcal{O}_S(V)}(\mathcal{O}_X(U), A_i) $$ Consider the schemes $T_i = \Spec(A_i)$. They form an inverse system of $V$-schemes over $I$ with transition morphisms $f_{ii'} : T_i \to T_{i'}$ induced by the $\mathcal{O}_S(V)$-algebra maps $\varphi_{i'i}$. Set $T := \Spec(A) = \lim_i T_i$. The formula above becomes in terms of morphism sets of schemes $$ \Mor_V(\lim_i T_i, U) = \colim_i \Mor_V(T_i, U). $$ We first observe that $\Mor_V(T_i, U) = \Mor_S(T_i, U)$ and $\Mor_V(T, U) = \Mor_S(T, U)$. Hence we have to show that $$ \Mor_S(\lim_i T_i, U) = \colim_i \Mor_S(T_i, U) $$ and we are given that $$ \Mor_S(\lim_i T_i, X) = \colim_i \Mor_S(T_i, X). $$ Hence it suffices to prove that given a morphism $g_i : T_i \to X$ over $S$ such that the composition $T \to T_i \to X$ ends up in $U$ there exists some $i' \geq i$ such that the composition $g_{i'} : T_{i'} \to T_i \to X$ ends up in $U$. Denote $Z_{i'} = g_{i'}^{-1}(X \setminus U)$. Assume each $Z_{i'}$ is nonempty to get a contradiction. By Lemma \ref{lemma-limit-closed-nonempty} there exists a point $t$ of $T$ which is mapped into $Z_{i'}$ for all $i' \geq i$. Such a point is not mapped into $U$. A contradiction. \medskip\noindent Finally, let us prove that (1) implies (3). Assume (1). Let an inverse directed system $(T_i, f_{ii'})$ of $S$-schemes be given. Assume the morphisms $f_{ii'}$ are affine and each $T_i$ is quasi-compact and quasi-separated as a scheme. Let $T = \lim_i T_i$. Denote $f_i : T \to T_i$ the projection morphisms. We have to show: \begin{enumerate} \item[(a)] Given morphisms $g_i, g'_i : T_i \to X$ over $S$ such that $g_i \circ f_i = g'_i \circ f_i$, then there exists an $i' \geq i$ such that $g_i \circ f_{i'i} = g'_i \circ f_{i'i}$. \item[(b)] Given any morphism $g : T \to X$ over $S$ there exists an $i \in I$ and a morphism $g_i : T_i \to X$ such that $g = f_i \circ g_i$. \end{enumerate} \noindent First let us prove the uniqueness part (a). Let $g_i, g'_i : T_i \to X$ be morphisms such that $g_i \circ f_i = g'_i \circ f_i$. For any $i' \geq i$ we set $g_{i'} = g_i \circ f_{i'i}$ and $g'_{i'} = g'_i \circ f_{i'i}$. We also set $g = g_i \circ f_i = g'_i \circ f_i$. Consider the morphism $(g_i, g'_i) : T_i \to X \times_S X$. Set $$ W = \bigcup\nolimits_{U \subset X\text{ affine open}, V \subset S\text{ affine open}, f(U) \subset V} U \times_V U. $$ This is an open in $X \times_S X$, with the property that the morphism $\Delta_{X/S}$ factors through a closed immersion into $W$, see the proof of Schemes, Lemma \ref{schemes-lemma-diagonal-immersion}. Note that the composition $(g_i, g'_i) \circ f_i : T \to X \times_S X$ is a morphism into $W$ because it factors through the diagonal by assumption. Set $Z_{i'} = (g_{i'}, g'_{i'})^{-1}(X \times_S X \setminus W)$. If each $Z_{i'}$ is nonempty, then by Lemma \ref{lemma-limit-closed-nonempty} there exists a point $t \in T$ which maps to $Z_{i'}$ for all $i' \geq i$. This is a contradiction with the fact that $T$ maps into $W$. Hence we may increase $i$ and assume that $(g_i, g'_i) : T_i \to X \times_S X$ is a morphism into $W$. By construction of $W$, and since $T_i$ is quasi-compact we can find a finite affine open covering $T_i = T_{1, i} \cup \ldots \cup T_{n, i}$ such that $(g_i, g'_i)|_{T_{j, i}}$ is a morphism into $U \times_V U$ for some pair $(U, V)$ as in the definition of $W$ above. Since it suffices to prove that $g_{i'}$ and $g'_{i'}$ agree on each of the $f_{i'i}^{-1}(T_{j, i})$ this reduces us to the affine case. The affine case follows from Algebra, Lemma \ref{algebra-lemma-characterize-finite-presentation} and the fact that the ring map $\mathcal{O}_S(V) \to \mathcal{O}_X(U)$ is of finite presentation (see Morphisms, Lemma \ref{morphisms-lemma-locally-finite-presentation-characterize}). \medskip\noindent Finally, we prove the existence part (b). Let $g : T \to X$ be a morphism of schemes over $S$. We can find a finite affine open covering $T = W_1 \cup \ldots \cup W_n$ such that for each $j \in \{1, \ldots, n\}$ there exist affine opens $U_j \subset X$ and $V_j \subset S$ with $f(U_j) \subset V_j$ and $g(W_j) \subset U_j$. By Lemmas \ref{lemma-descend-opens} and \ref{lemma-limit-affine} (after possibly shrinking $I$) we may assume that there exist affine open coverings $T_i = W_{1, i} \cup \ldots \cup W_{n, i}$ compatible with transition maps such that $W_j = \lim_i W_{j, i}$. We apply Algebra, Lemma \ref{algebra-lemma-characterize-finite-presentation} to the rings corresponding to the affine schemes $U_j$, $V_j$, $W_{j, i}$ and $W_j$ using that $\mathcal{O}_S(V_j) \to \mathcal{O}_X(U_j)$ is of finite presentation (see Morphisms, Lemma \ref{morphisms-lemma-locally-finite-presentation-characterize}). Thus we can find for each $j$ an index $i_j \in I$ and a morphism $g_{j, i_j} : W_{j, i_j} \to X$ such that $g_{j, i_j} \circ f_i|_{W_j} : W_j \to W_{j, i} \to X$ equals $g|_{W_j}$. By part (a) proved above, using the quasi-compactness of $W_{j_1, i} \cap W_{j_2, i}$ which follows as $T_i$ is quasi-separated, we can find an index $i' \in I$ larger than all $i_j$ such that $$ g_{j_1, i_{j_1}} \circ f_{i'i_{j_1}}|_{W_{j_1, i'} \cap W_{j_2, i'}} = g_{j_2, i_{j_2}} \circ f_{i'i_{j_2}}|_{W_{j_1, i'} \cap W_{j_2, i'}} $$ for all $j_1, j_2 \in \{1, \ldots, n\}$. Hence the morphisms $g_{j, i_j} \circ f_{i'i_j}|_{W_{j, i'}}$ glue to given the desired morphism $T_{i'} \to X$.