**Proof.**
Assume $f$ is closed. Let $y' \leadsto y$ in $Y$ and any $x'\in X$ with $f(x') = y'$ be given. Consider the closed subset $T = \overline{\{ x'\} }$ of $X$. Then $f(T) \subset Y$ is a closed subset, and $y' \in f(T)$. Hence also $y \in f(T)$. Hence $y = f(x)$ with $x \in T$, i.e., $x' \leadsto x$.

Assume $f$ is open, $X$ Noetherian, every irreducible closed subset of $X$ has a generic point, and $Y$ is Kolmogorov. Let $y' \leadsto y$ in $Y$ and any $x \in X$ with $f(x) = y$ be given. Consider $T = f^{-1}(\{ y'\} ) \subset X$. Take an open neighbourhood $x \in U \subset X$ of $x$. Then $f(U) \subset Y$ is open and $y \in f(U)$. Hence also $y' \in f(U)$. In other words, $T \cap U \not= \emptyset $. This proves that $x \in \overline{T}$. Since $X$ is Noetherian, $T$ is Noetherian (Lemma 5.9.2). Hence it has a decomposition $T = T_1 \cup \ldots \cup T_ n$ into irreducible components. Then correspondingly $\overline{T} = \overline{T_1} \cup \ldots \cup \overline{T_ n}$. By the above $x \in \overline{T_ i}$ for some $i$. By assumption there exists a generic point $x' \in \overline{T_ i}$, and we see that $x' \leadsto x$. As $x' \in \overline{T}$ we see that $f(x') \in \overline{\{ y'\} }$. Note that $f(\overline{T_ i}) = f(\overline{\{ x'\} }) \subset \overline{\{ f(x')\} }$. If $f(x') \not= y'$, then since $Y$ is Kolmogorov $f(x')$ is not a generic point of the irreducible closed subset $\overline{\{ y'\} }$ and the inclusion $\overline{\{ f(x')\} } \subset \overline{\{ y'\} }$ is strict, i.e., $y' \not\in f(\overline{T_ i})$. This contradicts the fact that $f(T_ i) = \{ y'\} $. Hence $f(x') = y'$ and we win.
$\square$

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