$Z^{A\amalg B}\cong Z^A\times Z^B$?

Question

Let $\mathcal{C}$ be a category with products, coproducts and exponentials and let $Z^{A\amalg B}$ be the exponential object for the coproduct $A\amalg B$. Is it true that $Z^{A\amalg B}$ is isomorphic to the product of $Z^A$ and $Z^B$?

Answer 1

Consider morphisms $X \to Z^{A+B}$ for an arbitrary object $X$.

• Under the exponential adjunction, these correspond with morphisms $(A+B) \times X \to Z$.
• Under the exponential adjunction again (using $P \times Q \cong Q \times P$), these correspond with morphisms $A+B \to Z^X$.
• By the universal property of the coproduct, these correspond with pairs of morphisms $A \to Z^X$ and $B \to Z^X$, i.e. elements of $\mathrm{Hom}(A,Z^X) \times \mathrm{Hom}(B,Z^X)$.
• Under the exponential adjunction (twice again), these correspond with pairs of morphisms $X \to Z^A$ and $X \to Z^B$.
• By the universal property of the product, these correspond with morphisms $X \to Z^A \times Z^B$.

More concisely the string of correspondences looks like: $$\begin{align} \mathrm{Hom}_{\mathcal{C}}(X, Z^{A+B}) & \cong \mathrm{Hom}_{\mathcal{C}}((A+B) \times X, Z) \\ & \cong \mathrm{Hom}_{\mathcal{C}}(X \times (A+B), Z) \\ & \cong \mathrm{Hom}_{\mathcal{C}}(A+B, Z^X) \\ & \cong \mathrm{Hom}_{\mathcal{C}}(A, Z^X) \times \mathrm{Hom}_{\mathcal{C}}(B, Z^X) \\ & \cong \mathrm{Hom}_{\mathcal{C}}(X \times A, Z) \times \mathrm{Hom}_{\mathcal{C}}(X \times B, Z) \\ & \cong \mathrm{Hom}_{\mathcal{C}}(A \times X, Z) \times \mathrm{Hom}_{\mathcal{C}}(B \times X, Z) \\ & \cong \mathrm{Hom}_{\mathcal{C}}(X, Z^A) \times \mathrm{Hom}_{\mathcal{C}}(X, Z^B) \\ & \cong \mathrm{Hom}_{\mathcal{C}}(X, Z^A \times Z^B) \end{align}$$

You need to prove that all of these correspondences are natural in $X$ and $Z$ (for fixed $A,B$), and hence $Z^{A+B} \cong Z^A \times Z^B$ by the Yoneda lemma.

P.S. I write $+$ where you wrote $\amalg$; force of habit.