A category with products is said to have exponentials if for all objects $x, y$ there exists an object $y^x$ equipped with an arrow $e\colon x\times y^x\to y$ such that for all objects $z$ and all arrows $f\colon x\times z\to y$ there is a unique arrow $\bar{f}\colon z\to y^x$ satisfying $e\circ (id_x\times\bar{f})=f$.
I see that if a category has exponentials, then $f\mapsto \bar{f}$ is a natural isomorphism between $hom(x\times z, y)$ and $hom(z, y^x)$ with inverse $\bar{f}\mapsto id_x\times\bar{f}$. Hence the functor $x\times (-)$ is left adjoint to $(-)^x$.
I am wondering about the converse: if $C$ is a category with products such that $x\times (-)$ has a right adjoint, does it follow that $C$ has exponentials?
In particular, if we just assume that $x\times (-)$ has a right adjoint, how do we equip $y^x$ with the arrow $e\colon x\times y^x\to y$. Also, how do we deduce that the equation $e\circ (id_x\times\bar{f})=f$ holds precisely?
Somehow the existence of a right adjoint of $x\times (-)$ feels weaker and more abstract than the universal property definition of a category having exponentials given above.
I suppose one needs AC to choose an object $y^x$ for each $x$ and $y$.
Accepting this, one gets the arrow $e$ from the formalism of units/counits in adjunctions. If $F$ is a right adjoint of $x\times(-)$ then naturally, $$\text{hom}(a,Fy)\cong\text{hom}(x\times a,y).$$ Take $a=Fy$. Then $$\text{hom}(Fy,Fy)\cong\text{hom}(x\times Fy,y).$$ The identity on the left maps to a homomorphism $e:x\times Fy\to y$ on the right. We are denoting $Fy$ as $y^x$, and this $e:x\times y^x\to y$ is the exponential map.