Proof.
We have seen the equivalence of (1) and (2) and the fact that these imply (3) in Lemma 32.15.4. Thus it suffices to prove that (3) implies (2). Observe that if condition (3) holds for $f : X \to Y$, then condition (3) holds for $1 \times f : \mathbf{A}^ n \times X \to \mathbf{A}^ n \times Y$ (see argument in the proof of Lemma 32.15.4). Hence it suffices to show that (3) implies that $f$ is closed.
Reduction to the case where $Y$ and $S$ are affine; we suggest skipping this paragraph. Let $S' \subset S$ be an affine open and let $Y' \subset Y$ be an affine open mapping into $S'$. Set $X' = f^{-1}(Y')$. Then we claim that the restriction $f' : X' \to Y'$ of $f$ viewed as a morphism of schemes over $S'$ has property (3) also. We omit the details. Now if we can prove that $f'$ is closed for all choices of $S'$ and $Y'$, then it follows that $f$ is closed. This reduces us to the case discussed in the next paragraph.
Assume $S$ and $Y$ affine. Let $Z \subset X$ be a closed subset. We may and do view $Z$ as a reduced closed subscheme of $X$. We have to show that $E = f(Z)$ is closed. Pick $y \in Y$ a closed point contained in the closure of $f(Z)$. It suffices to show $y \in E$. We assume $y \not\in E$ to get a contradiction. The image $s \in S$ of $y$ is a finite type point of $S$, see Morphisms, Lemma 29.16.5. Recall that $E$ is constructible (Morphisms, Lemma 29.22.2). Consider the intersection $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \cap E$. This is a constructible subset of the spectrum (Morphisms, Lemma 29.22.1) which doesn't contain the closed point. Since the punctured spectrum $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \setminus \{ y\} $ is Jacobson (Morphisms, Lemma 29.16.10), we find a closed point $t \in \mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y}) \setminus \{ y\} $ with $t \in E$ (see Topology, Lemma 5.18.5). In other words, $t \in E$ is a point of $Y$ which has an immediate specialization $t \leadsto y$. As $t \in E$ the scheme theoretic fibre $Z_ t$ is nonempty. Choose a closed point $x \in Z_ t$. In particular we have $[\kappa (x) : \kappa (t)] < \infty $ by the Hilbert Nullstellensatz (Morphisms, Lemma 29.20.3).
Denote $T = \overline{\{ t\} } \subset Y$ the integral closed subscheme whose underlying topological space is as indicated (Schemes, Definition 26.12.5). Then $t \in T$ is the generic point. Denote $C \to T$ the normalization of $T$ in $\kappa (x)$, see Morphisms, Section 29.53 (more precisely, $C \to T$ is the normalization of $T$ in $x$ where we view $x = \mathop{\mathrm{Spec}}(\kappa (x)) \to T$ as a scheme over $T$). Since $S$ is a Nagata scheme, so is $T$ (Morphisms, Lemma 29.18.1). Hence we see that $C \to T$ is finite (Morphisms, Lemma 29.53.14). As $t$ is in the image we see that $C \to T$ is surjective (because the image is closed and $T$ is the closure of $t$ in $Y$). Choose a point $c \in C$ mapping to $y \in T$. Since $y$ is a closed point of $T$ we see that $c$ is a closed point of $C$. Since $\dim (\mathcal{O}_{T, y}) = 1$ we see that $\dim (\mathcal{O}_{C, c}) = 1$ (the dimension is at least $1$ as $c$ is not the generic point of $C$ and at most $1$ as $C \to T$ is finite). As the function field of $C$ is $\kappa (x)$ and as $x$ is a point of $X$, we have a $Y$-rational map from $C$ to $X$ (see for example Morphisms, Lemma 29.49.2). Let $C \supset U \to X$ be a representative (in particular $U$ is nonempty). We may assume $c \not\in U$ (replace $U$ by $U \setminus \{ c\} $). Since $c$ is a closed point of codimension $1$ in the integral scheme $C$ we have $C = U \amalg \{ c\} \amalg \Sigma $ for some proper closed subset $\Sigma \subset C$. After replacing $C$ by $C \setminus \Sigma $ we have constructed a commutative diagram as in part (3). By the 2nd footnote in the statement of the lemma, the existence of the dotted arrow produces an extension of the rational map to all of $C$ and we get the contradiction because the image of $c$ will be a point of $Z$ mapping to $y$.
$\square$
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