# Equivalence between two definitions of a category having exponential objects

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.

• The two definitions are equivalent ; the map $x\times y^x\to y$ is just the $y$ component of the counit, so $e\circ (id_x\times\bar{f})=f$ comes from the usual construction of the bijection in terms of unit and counit. Commented Aug 15, 2020 at 13:31

## 1 Answer

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.