We prove (1) and (2) and we omit the proof of (3) and (4) and the final statement. Consider the category of $A$-algebras of the form $$ B = A[x_1, \ldots, x_n]/J $$ where $J$ is a finitely generated ideal such that $A \to B$ defines a universal homeomorphism on spectra. We claim this category is directed (Categories, Definition \ref{categories-definition-directed}). Namely, given $$ B = A[x_1, \ldots, x_n]/J \quad\text{and}\quad B' = A[x_1, \ldots, x_{n'}]/J' $$ then we can consider $$ B'' = A[x_1, \ldots, x_{n + n'}]/J'' $$ where $J''$ is generated by the elements of $J$ and the elements $f(x_{n + 1}, \ldots, x_{n + n'})$ where $f \in J'$. Then we have $A$-algebra homomorphisms $B \to B''$ and $B' \to B''$ which induce an isomorphism $B \otimes_A B' \to B''$. It follows from Lemmas \ref{lemma-base-change-universal-homeomorphism} and \ref{lemma-composition-universal-homeomorphism} that $\Spec(B'') \to \Spec(A)$ is a universal homeomorphism and hence $A \to B''$ is in our category. Finally, given $\varphi, \varphi' : B \to B'$ in our category with $B$ as displayed above, then we consider the quotient $B''$ of $B'$ by the ideal generated by $\varphi(x_i) - \varphi'(x_i)$, $i = 1, \ldots, n$. Since $\Spec(B') = \Spec(B)$ we see that $\Spec(B'') \to \Spec(B')$ is a bijective closed immersion hence a universal homeomorphism. Thus $B''$ is in our category and $\varphi, \varphi'$ are equalized by $B' \to B''$. This completes the proof of our claim. We set $$ A^{awn} = \colim B $$ where the colimit is over the category just described. Observe that $A \to A^{awn}$ induces a universal homeomorphism on spectra by Lemma \ref{lemma-colimit-inherits} (this is where we use the category is directed). \medskip\noindent Given a ring map $A \to B$ of finite presentation inducing a universal homeomorphism on spectra, we get a canonical map $B \to A^{awn}$ by the very construction of $A^{awn}$. Since every $A \to B$ as in (1) is a filtered colimit of $A \to B$ as in (1) of finite presentation (Lemma \ref{lemma-universal-homeo-limit}), we see that $A \to A^{awn}$ is final in the category (1). \medskip\noindent Let $x, y \in A^{awn}$ be elements such that $x^3 = y^2$. Then $A^{awn} \to A^{awn}[t]/(t^2 - x, t^3 - y)$ induces a universal homeomorphism on spectra by Lemma \ref{lemma-make-universal-homeo}. Thus $A \to A^{awn}[t]/(t^2 - x, t^3 - y)$ is in the category (1) and we obtain a unique $A$-algebra map $A^{awn}[t]/(t^2 - x, t^3 - y) \to A^{awn}$. The image $a \in A^{awn}$ of $t$ is therefore the unique element such that $a^2 = x$ and $a^3 = y$ in $A^{awn}$. In exactly the same manner, given a prime $p$ and $x, y \in A^{awn}$ with $p^px = y^p$ we find a unique $a \in A^{awn}$ with $a^p = x$ and $pq = y$. Thus $A^{awn}$ is absolutely weakly normal by definition. \medskip\noindent Finally, let $A \to B$ be in the category (1) with $B$ absolutely weakly normal. Since $A^{awn} \to B^{awn}$ is a universal homeomorphism and  since $A^{awn}$ is reduced (Lemma \ref{lemma-seminormal-reduced}) we find $A^{awn} \subset B^{awn}$ (see Algebra, Lemma \ref{algebra-lemma-image-dense-generic-points}). If this inclusion is not an equality, then Lemma \ref{lemma-pth-power-and-multiple} implies there is an element $b \in B^{awn}$, $b \not \in A^{awn}$ such that either $b^2, b^3 \in A^{awn}$ or $pb, b^p \in A^{awn}$ for some prime number $p$. However, by the existence and uniqueness in Definition \ref{definition-seminormal-ring} this forces $b \in A^{awn}$ and hence we obtain the contradiction that finishes the proof.