How do you express coexponentials?

Question

If a coexponential object in $C$ is an exponential object in $C^{\text{op}}$,
that is, an object that behaves just like an exponential object if $C^\text{op}$ were to be treated divorced from $C$, how would this carry over to $C$, and how would it be written down?

After all, I would contend that a coproduct $A + B$ is characterized not by its behaviour in the opposite category, but by:
If some $f_A : K \rightarrow A$ or $f_B : K \rightarrow B$ are in $C$, then for $A+B$, $C$ contains $f : K \rightarrow A + B$.

Is a similar expansion possible for coexponentials,
and how do you write them? $A_K$, $\log_K(A)$, $\root{K}\of{A}$, $ ^KA$?

Answer 1

$\require{AMScd}$

A quick answer will be that given any object $A$ in $\mathbf{C}$ such that all the binary products $A \times B$ exist we get a functor $A \times - : \mathbf{C} \to \mathbf{C}$. If this functor has a right adjoint $A \times - \dashv (-)^A$ then $B^A$ is an exponential object. Conversely if all the exponential object $B^A$ exist then $(-)^A$ is the right adjoint of $A \times -$.

Dually coexponential objects $B_A$ correspond to a left adjoint $(-)_A \dashv A \sqcup -$. You can also get the universal property of a coexponential object directly for the universal property of an exponential object by reversing all the arrows involved. As far as I am aware of there is no standard notation for coexponential objects.

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Let me spill out some of the details. From this we should also get the universal property of a coexponential object.

The way you may have been introduced to exponential objects is through their universal property.

It would be an object $C^A$ equipped with a map $ev_{A,C} : A \times C^A \to C$ such that for any morphism $f : A \times B \to C$ there is a unique morphism $\lambda(f) : B \to C^A$ with $f = ev_{A,C} \circ (id_A \times \lambda(f))$.

Saying that there is an adjunction $A \times - \dashv (-)^A$ means that there are natural isomorphisms $\mathbf{C}(A \times B, C) \simeq \mathbf{C}(B,C^A)$ where $\mathbf{C}(X,Y)$ is the set of morphisms from $X$ to $Y$.

We can show that these are equivalent. I give a sketch of a proof. This is also mentioned in exponential object and cartesian closed category.

Suppose that for any $A$ we have an exponential object $C^A$. Then we get a function $\lambda(-) : \mathbf{C}(A \times B, C) \to \mathbf{C}(B,C^A)$ with inverse $ev_{A,B} \circ (id_A \times -)$. I will let you check that this isomorphism is natural.

On the other hand suppose that we have natural isomorphisms $\lambda : \mathbf{C}(A \times B, C) \to \mathbf{C}(B,C^A)$ then we get $ev_{A,C} = \lambda^{-1}(id_{C^A})$. I will also let you check that these maps give the universal property of the exponential object.

Now, recall that the product in $\mathbf{C}^{op}$ is a coproduct in $\mathbf{C}$, so we will note it $\sqcup$. Suppose then that for any object $A$ $A \sqcup -$ has a left adjoint $(-)_A$ in $\mathbf{C}^{op}$. This means that $\mathbf{C}^{op}(A \sqcup B, C) \simeq \mathbf{C}^{op}(B,C_A)$. Dually, $\mathbf{C}(C, A \sqcup B) \simeq \mathbf{C}(C_A,B)$. And naturality will follow from naturality in the opposite category.

In other words, $(-)_A$ is the left adjoint of $A \sqcup -$ in $\mathcal{C}$. You should be able to turn this around to get a universal property for the coexponential.

A coexponential is an object $C_A$ equipped with a morphism $coev_{A,C} : C \to A \sqcup C_A$ such that for any morphism $f : C \to A \sqcup B$ there is a unique map $co\lambda(f) : C_A \to B$ with $f = (id_A \sqcup co\lambda(f)) \circ coev_{A,C}$.

Answer 2

Well, the exponential object $[X,Y]$ can be described as the terminal object of the following category:

• Objects are pairs $(F,\,\mathrm{eval}_F:F \times X \rightarrow Y)$
• A morphism $\varphi : (F,\,\mathrm{eval}_F:F \times X \rightarrow Y) \rightarrow (G,\,\mathrm{eval}_G: G \times X \rightarrow Y)$ is a morphism $\varphi: F \rightarrow G$ satisfying $\mathrm{eval}_G \circ (\varphi \times X) = \mathrm{eval}_F$.

Thus the coexponential (as you define it), when it exists, can be described as the initial object of the following category:

• Objects are pairs $(F,\,\mathrm{coeval}_F:X \rightarrow F \sqcup Y)$
• Morphisms are required to satisfy $(\varphi \sqcup Y) \circ \mathrm{coeval}_F = \mathrm{coeval}_G$.

I don't know of any non-thin categories with these objects, but thin categories that have such "coexponential objects" abound. In particular, the lattice-theoretic decategorification of this notion is pretty important. Given that $X$ and $Y$ are elements of a preordered set, the coexponential of $X$ and $Y$, when it exists, is the least $F$ satisfying $X \leq F \vee Y$. Such $F$'s always exist in powerset lattices, and are commonly denoted $X \setminus Y$.