 Solution

A regular category has finite limits, regular-epimorphism--monomorphism image factorizations, and regular epimorphisms stable under pullback. A relation $A⇝B$ in $\mathbf{Rel}(\mathcal{C})$ is a subobject of $A\times B$; composition forms the pullback over the middle object and then takes its image.

Suppose $R:A⇝B$ has a right adjoint relation $S$, so $1_A\le SR$ and $RS\le 1_B$. In the internal regular logic, the first inequality says that for every $a$ there is a $b$ with $R(a,b)$ and $S(b,a)$. If also $R(a,b^{\prime })$, the second inequality forces $b=b^{\prime }$. Thus $R$ is total and single-valued. Categorically, if $R\hookrightarrow A\times B$ has projections $p:R\to A$ and $q:R\to B$, totality makes $p$ a regular epimorphism and single-valuedness makes it a monomorphism. Hence $p$ is an isomorphism and $R$ is the graph of a morphism as a relation $q{p}^{-1}:A\to B$. Conversely, the graph of any morphism is left adjoint to its converse relation, as the two required inequalities follow directly from equality. This proves the characterization.

Let $L$ be a frame. Composition in the category of matrices valued in a frame isIf $f⊣g$, the diagonal part of $1_A\le gf$ impliesso every row of $f$ joins to $1$. For $b\ne b^{\prime }$, distribute $f(a,b)\wedge f(a,b^{\prime })$ over the displayed join. Every term vanishes by $fg\le 1_B$, first using the factor with column $b$ and then the one with column $b^{\prime }$. HenceConversely, if the rows of $f$ join to $1$ and have pairwise disjoint entries, define $g(b,a)=f(a,b)$. Thenwhile $(fg)(b,b^{\prime })=0$ for $b\ne b^{\prime }$. Thus $1_A\le gf$ and $fg\le 1_B$, proving the stated criterion.

Finally take $L=\Omega (X)$ for a connected topological space $X$. For fixed $a$, the opens $f(a,b)$ are pairwise disjoint and cover $X$. Connectedness forces exactly one of them to be $X$ and all the others to be empty. Hence a left adjoint matrix determines a unique function $\phi :A\to B$ by $f(a,\phi (a))=X$. Conversely every function gives this matrix, and matrix composition agrees with function composition. The left adjoints in $\mathbf{Mat}(\Omega (X))$ therefore form a category isomorphic to $\mathbf{Set}$.

Solved by gpt-5.6-sol high.