Lemma 29.58.2. Let $f : Y \to X$ be a morphism of schemes. Let $E \subset X$. Assume $X$ is locally Noetherian, there are no nontrivial specializations among the elements of $E$, $Y$ is reduced, and $f(Y) \subset E$. Then $f$ factors through $\coprod _{x \in E} x \to X$.

**Proof.**
When $E$ is a singleton this follows from Lemma 29.58.1. If $E$ is finite, then $E$ (with the induced topology of $X$) is a finite discrete space by our assumption on specializations. Hence this case reduces to the singleton case. In general, there is a reduction to the case where $X$ and $Y$ are affine schemes. Say $f : Y \to X$ corresponds to the ring map $\varphi : A \to B$. Denote $A' \subset B$ the image of $\varphi $. Let $E' \subset \mathop{\mathrm{Spec}}(A') \subset \mathop{\mathrm{Spec}}(A)$ be the set of minimal primes of $A'$. By Algebra, Lemma 10.30.5 the set $E'$ is contained in the image of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A') \subset \mathop{\mathrm{Spec}}(A)$. We conclude that $E' \subset E$. Since $A'$ is Noetherian we have $E'$ is finite by Algebra, Lemma 10.31.6. Since any other point in the image of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is a specialization of an element of $E'$ and in $E$, we conclude that the image is contained in $E'$ (by our assumption on specializations between points of $E$). Thus we reduce to the case where $E$ is finite which we dealt with above.
$\square$

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