W-split coequalizers

Question

The following snippet is from Adamek, Rosicky:Algebra and local presentability,how algebraic are.

It is unclear to me the end of Example 5.1:

Since $e$ is the coequalizer of $\bar{u}_1,\bar{u}_2$ in $\mathbf{Pos}$, we conclude that $W$ does not preserve $W-$split coequalizers.

Why?

The snippet:

Answer 1

$e$ is the coequalizer in $\mathbf{Pos}$, but it can't be in $\mathbf{Alg}$ : $L/E$ is not algebraic !

So if $f$ is the coequalizer of $\overline{u_1,u_2}$ in $\mathbf{Alg}$, $W(f) \neq e$ (or rather : the codomain of $W(f)$ is not isomorphic to that of $e$ (as it is an algebraic lattice, as opposed to that of $e$) so $W(f)$ is not a coequalizer in $\mathbf{Pos}$) so $W$ does not preserve the equalizer in question.

Answer 2

There are a mistake and a few unexplicated points in this passage. First, we need to take the domain of the $\bar u_i$ to be $\mathrm{Ind} E$, not $\mathrm{Ind} L$. The existence of $\bar u_1$ and $\bar u_2$ is then exactly the universal property of $\mathrm{Ind} E$: a map $u$ out of $E$ into a poset $L$ with directed joins extends uniquely to a directed join-preserving map out of $\mathrm{Ind} E$ by sending a directed downward-closed set $S\subset E$ to the join of the directed set $u(S)$ in $L$.

The assumption that $e$ preserve meets and directed joins is equivalent to the assumption that $E$ is closed in $L\times L$ under meets and directed joins. In particular, $E$ has meets, so $\mathrm{Ind} E$ is an algebraic lattice in which the representable downsets $\downarrow x=\{y:y\leq x\}$ are compact elements.

Now, a key point: $\bar u_1$ and $\bar u_2$ generate precisely the congruence $E$ on the poset $L$. Indeed, for $S\in\mathrm{Ind}E$ we have $\bar u_1(S)=\vee_{s\in S}u_1(s)=u_1(\vee_{s\in S} s)$. So if $x=u_1(s)$ and $y=u_2(s)$, then $x=\bar u_1(\downarrow s)$ and $y=\bar u_2(\downarrow s)$. Conversely if $x=\bar u_1(S)$ and $y=\bar u_2(S)$, then $x=u_1(\vee S)$ and $y=u_2(\vee S)$. This explains the claim that $e$ is the coequalizer of $\bar u_1$ and $\bar u_2$ in $\mathbf{Pos}$.

However, $e$ cannot be the coequalizer of $\bar u_1$ and $\bar u_2$ in $\mathbf{Alg}$, since $L/E$ is non-algebraic by assumption! As to why $\bar u_1,\bar u_2$ is $W$-split, it is a contractible pair. And this follows from the fact that $u_1,u_2:E\to L$ is a contractible pair, with the splitting $\ell\mapsto (\ell,\wedge_{\ell'E\ell} \ell')$. To show that $(\bar u_1,\bar u_2)$ is contractible, we simply compose this splitting with the left adjoint of the join map $\mathrm{Ind}E\to E$, which is a split mono.

So, $W$ does not create $W$-split coequalizers, which is all that really needs to be shown. To show that $W$ does not even preserve $W$-contractible coequalizers, it suffices to know that $\mathbf{Alg}$ admits coequalizers. The easiest way I can think of to show the latter is to use Gabriel-Ulmer duality, which says that $\mathbf{Alg}$ is contravariantly equivalent to the category of posets admitting finite meets.