= Solution
A <regular category> has finite limits, regular-epimorphism--monomorphism image factorizations, and regular epimorphisms stable under pullback. A relation $A\rightsquigarrow B$ 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 $R:A\rightsquigarrow B$ has a right adjoint relation $S$, so $1_A\leq SR$ and $RS\leq1_B$. 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 $R(a,b')$, the second inequality forces $b=b'$. 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> $qp^{-1}:A\to B$. 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> is
$$
(gf)(a,c)=\bigvee_{b\in B}f(a,b)\wedge g(b,c).
$$
If $f\dashv g$, the diagonal part of $1_A\leq gf$ implies
$$
1=\bigvee_b f(a,b)\wedge g(b,a)\leq\bigvee_bf(a,b),
$$
so every row of $f$ joins to $1$. For $b\ne b'$, distribute $f(a,b)\wedge f(a,b')$ over the displayed join. Every term vanishes by $fg\leq1_B$, first using the factor with column $b$ and then the one with column $b'$. Hence
$$
f(a,b)\wedge f(a,b')=0.
$$
Conversely, if the rows of $f$ join to $1$ and have pairwise disjoint entries, define $g(b,a)=f(a,b)$. Then
$$
(gf)(a,a)=\bigvee_bf(a,b)=1,
$$
while $(fg)(b,b')=0$ for $b\ne b'$. Thus $1_A\leq gf$ and $fg\leq1_B$, 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 $\varphi:A\to B$ by $f(a,\varphi(a))=X$. 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 $\mathbf{Set}$.
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