Does the contravariant exponential functor have a left adjoint ? And if so what is it ?
To elaborate, I know that in many categories, like $\textbf {Set}$ for example, the covariant exponential functor $\left(\_ \right)^A$ has a left adjoint (which is the product functor $\left(\_ \right)\times A$). However, given an object $B,$ there is also a contravariant functor $B^{\left(\_ \right)}$ associated with exponentiation. Here $B^{\left(\_ \right)}$ sends an object $A$ to $B^A.$ Also, $B^{\left(\_ \right)}$ sends an arrow $g: A\rightarrow A'$ to the arrow $B^f: B^{A'} \rightarrow B^A,$ where $B^f$ is the exponential transpose of $e \left( 1_{B^{A'}} \times g \right).$
So my question is can the functor $B^{\left(\_ \right)}$ have a left adjoint ? And if so what is it ? Or to be more specific, I could ask, does $B^{\left(\_ \right)}$ have a left adjoint when we are working in $\textbf {Set},$ or some more general Cartesian closed category ?
Basically, I am trying to understand the nature of the contravariant exponential functor. If it helps, I would also like to know about its right adjoint (if it exists).
Let $\cal C$ be your category. Contravariant functors confuse me, so I'll think of $B^{(-)}$ as a functor $\cal C^{\textrm{op}}\to\cal C$. So I presume we want a functor $F:\cal C\to\cal C^{\textrm{op}}$ with $${\text{Hom}}_{\cal C}(Y,B^X)\cong{\text{Hom}}_{\cal C^{\textrm{op}}}(F(Y),X),$$ naturally, that is $${\text{Hom}}_{\cal C}(Y,B^X)\cong{\text{Hom}}_{\cal C}(X,F(Y)).$$ I reckon that $F(Y)=B^Y$ works here, certainly in the case of the category of sets, as $${\text{Hom}}_{\cal C}(Y,B^X)\cong{\text{Hom}}_{\cal C}(Y\times X,B) \cong{\text{Hom}}_{\cal C}(X\times Y,B)\cong{\text{Hom}}_{\cal C}(X,B^Y).$$