Understanding a lemma in proof of Valuative Criteria of Separatedness

Question

I am reading this document here by Brian Osserman concerning the Valuative Criteria for properness and separatedness. In particular, I am trying to understand the second half of the proof Proposition 2.5. Here is the proposition:

Proposition $\mathbf{2.5}$. Let $X$ be a scheme, and $x,x'\in X$ with $x$ specializing to $x'$. Then there exists a valuation ring $A$ and a morphism $\operatorname{Spec}A\to X$ with the generic point of $\operatorname{Spec}A$ mapping to $x$, and the closed point of $\operatorname{Spec} A$ mapping to $x'$.

More generally, if $f:X\to Y$ is any morphism, and we have $x\in X$, and $y'\in Y$ a specialization of $f(x)$, then there exists a valuation ring $A$, with fraction field $K$, and a commutative diagram $$\require{AMScd}\begin{CD} \text{Spec }K @>>> X\\@VVV @VVfV\\ \text{Spec }A @>>> Y \end{CD}$$ such that the image of $\operatorname{Spec}K$ is $x$, and the generic and closed points of $\operatorname{Spec}A$ map to $f(x)$ and $y$ respectively.

Now I can understand most of the proof (which is later in the document) except for the following reduction:

Proof of Proposition 2.5. Note that the first statement follows from the second, by taking $f$ to be the identity map. For the second statement, by replacing $Y$ with the closure of $f(x)$ (with the reduced induced subscheme structure), it is clear that we may assume $Y$ is integral, and $f(x)$ is the generic point of $Y$. Then, by Proposition 3.1, there exists a valuation ring $A'$ of $K(Y)$ dominating

My question is: Why can we assume that $Y$ is integral and that the $f(x)$ is the generic point of $Y$? It is clear to me that the map $f$ when restricted to $\overline{\{x\}}$ has image landing in $\{\overline{f(x)}\}$ (this is just continuity of $f$).

Answer 1

If $x \in X$, then $\overline{\{x\}}$ is a closed irreducible subset of $X$, hence can be regarded as a reduced irreducible, i.e. integral subscheme of $X$. Besides, if $f : X \to Y$ is a morphism of schemes, it induces not just a continuous map $\overline{\{x\}} \to \overline{\{f(x)\}}$, but also a dominant morphism of integral schemes $f'$. Locally, this comes down to: If $A \to B$ is a ring homomorphism and $\mathfrak{q} \subseteq B$ is a prime ideal, then for $\mathfrak{p} :=A \cap \mathfrak{q}$ we have an injective homomorphism $A/\mathfrak{p} \to B/\mathfrak{q}$. If the claim is proven for $f'$, it also follows for $f$, as is shown in the commutative diagram

$$\begin{array}{c} \mathrm{Spec}(K) & \rightarrow & \overline{\{x\}} & \rightarrow & X \\ \downarrow & & \downarrow && \downarrow \\ \mathrm{Spec}(R) & \rightarrow & \overline{\{f(x)\}} & \rightarrow & Y. \end{array}$$