Let $\mathcal{C}$ be a cartesian closed category.
I've read that you can understand the notion of a Boolean algebra internally to $\mathcal{C}$ as an object $B$, with morphism $\wedge: B\times B \to B$ and $\neg:B\to B$ with commuting diagrams corresponding to the usual axioms for Boolean algebras.
Apparently it is possible to state the completeness of $B$ using adjoints in something like the following way:
Let $k = \lambda \pi_1: B \to B^I$, the exponential transpose of $\pi_1: B\times I \to B$. Then the $I$ indexed conjunction, $\bigwedge_I: (B^I \to B)$ can be defined as a right adjoint of $k$.
But what I don't understand about this definition is that $k$ and $\bigwedge_I$ are not functors, or functions on a partial order so I don't know how to apply the usual notion of an adjunction/galois connection here (which is given by pointwise inequalities). I could try to spell it out in terms of elements $b:1 \to B$, but for all I've said $B$ has no elements. What is the "standard" way of making this definition precise in the categorical setting?
Since $\mathcal{C}$ is cartesian closed, it behaves enough like the category $\mathbf{Set}$ that the usual construction (in $\mathbf{Set}$) of $I$-ary conjunction $\bigwedge_I : B^I \to B$ as a right adjoint to $B \to B^I$, with $B$ viewed as a posetal category, goes through in $\mathcal{C}$ just by internalising the definitions.
More precisely: an internal boolean algebra $B$ in a cartesian closed category $\mathcal{C}$ has the structure of an internal category in $\mathcal{C}$ — to see this, just 'internalise' the usual proof that posets are categories. The morphism $k : B \to B^I$ extends to an internal functor in $\mathcal{C}$, i.e. a $1$-cell in the $2$-category $\mathbf{Cat}(\mathcal{C})$. This internal functor has an internal right adjoint $\bigwedge_I$, whose underlying morphism in $\mathcal{C}$ is the morphism $B^I \to B$ you desire.