Fix a commutative ring $R$. There's a contravariant functor from finite sets to finite $R$-algebras sending $X$ to $R^X$. Viewed as a covariant functor $\text{set}^{op}\to R\text{-alg}$, this functor preserves finite limits and colimits; this is easy to verify directly by checking finite products ($R^{\coprod_i X_i} \cong \prod_i R^{X_i}$) and coproducts ($R^{\prod_i X_i}\cong \bigotimes_i R^{X_i}$), as well as equalizers and coequalizers. I'd love to just cite this fact instead of (re)proving it in a paper I'm writing.
The preserving of limits comes from the fact that the contravariant functors $X\mapsto R^X$ and $A\mapsto Hom_{R\text{-alg}}(A,R)$ are "adjoint on the right":
$Hom_{R\text{-alg}}(A,R^X) \cong Hom_{R\text{-alg}}(A,R)^X \cong Hom_{\text{set}^{op}}(Hom_{R\text{-alg}}(A,R),X).$
If $R$ is Noetherian, then $X\mapsto R^X$ also has a right adjoint: an $R$-algebra homomorphism $R^X\to A$ corresponds to a complete set of orthogonal idempotents in $A$ indexed by $X$. If $A$ is finite (or even finite type) over $R$ then $Spec(A)$ is the topological disjoint union of its finitely many connected components, so a complete set of orthogonal idempotents is just a function $\pi_0(Spec(A))\to X$. Thus
$Hom_{R\text{-alg}}(R^X, A)\cong Hom_{\text{set}^{op}}(X, \pi_0(Spec(A)).$
But even with no extra hypotheses on $R$, the functor $X\mapsto R^X$ sends finite limits to colimits anyway. Does anyone know a reference for this, or something close to a one-liner proof so I can avoid a tedious nuts-and-bolts argument?