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An idempotent morphism satisfies $e^2=e$ and splits if $e=ir$ with $ri=1$. Form the relative Karoubi envelope $\mathcal{C}[\check{E}]$: its objects are $(A,e)$ for $e\in E$, andThe functor $A\mapsto (A,1_A)$ is full and faithful. Each $e\in E$ splits through $(A,e)$ using $e$ in both directions. If a functor $F:\mathcal{C}\to \mathcal{D}$ comes with splittings $F(e)=i_e{r}_e$, define its extension by $(A,e)\mapsto$ the splitting object and $u\mapsto r_{f}F(u)i_e$. This gives the required factorization, unique up to the unique compatible natural isomorphism.

A preorder $R$ is reflexive and transitive, so $R\circ R=R$: transitivity gives one inclusion and reflexivity the other. If it is an equivalence relation, the quotient relation $q:X\to X/R$ satisfies $q^{\mathsf{op}}q=R$ and $q{q}^{\mathsf{op}}=1$, so it splits in category of relations $\mathbf{Rel}$. Conversely, writing a split preorder as $R=ir$, $ri=1$ and using reflexivity and transitivity shows $R=R^{\mathsf{op}}$; hence it is an equivalence relation.

Let $B$ be the set of join-inaccessible elements of a completely algebraic lattice $A$, and putThen $ri=1_A$, while $r(S)\le a$ exactly when $S\subseteq i(a)$, so $r⊣i$. Join-inaccessibility gives $i({\bigvee }_j{a}_j)={\bigcup }_{j}i(a_j)$, so $i$ preserves all joins. Conversely, given such $r⊣i$ with $ri=1$, each $r\{b\}$ is join-inaccessible: if $r\{b\}\le \bigvee S$, adjunction and preservation of joins put $b$ in some $i(s)$, whence $r\{b\}\le s$. Finally,so $A$ is completely algebraic.

Under the supplied full embedding $\mathbf{Rel}\to \mathbf{CSLat}$, a set goes to its power-set lattice and a preorder goes to the associated idempotent. Splitting those idempotents gives exactly the retracts of power sets by adjoint join maps characterized above. Therefore $\mathbf{Rel}[\check{E}]$ is equivalent to the full subcategory of complete join-semilattices consisting of completely algebraic lattices.

Solved by gpt-5.6-sol high.