**Proof.**
Choose $U$, $N'$ as in the definition of a pure spreadout. Any element $x' \in N'$ has a content ideal in $A$ because $N'$ is $A$-projective (this can easily be seen directly, but it also follows from More on Algebra, Lemma 15.24.4 and Algebra, Example 10.91.1). Since $N' \to N$ is $A$-universally injective, we see that the image $x \in N$ of any $x' \in N'$ has a content ideal in $A$ (it is the same as the content ideal of $x'$). For a general $x \in N$ we choose $s \in S$ such that $s x$ is in the image of $N' \to N$ and we use that $x$ and $sx$ have the same content ideal.

Let $u : N \to M$ be as in (2). To show that $u$ is $A$-universally injective, we may replace $A$ by a localization at a maximal ideal (small detail omitted). Assume $A$ is local with maximal ideal $\mathfrak m$. Pick $s \in S$ and consider the composition

\[ N' \to N \xrightarrow {1/s} N \xrightarrow {u} M \]

Each of these maps is injective modulo $\mathfrak m$, hence the composition is $A$-universally injective by Lemma 38.7.5. Since $N = \mathop{\mathrm{colim}}\nolimits _{s \in S} (1/s)N'$ we conclude that $u$ is $A$-inversally injective as a colimit of universally injective maps.
$\square$

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