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Let $\mathcal{C}$ have products, $A, B \in \mathcal{C}_0$, ${\times}\colon \mathcal{C}^2\to \mathcal{C}$ be the functor that sends objects to their product. Then the induced product ${\boxtimes}\colon (\mathcal{C}/A, \mathcal{C}/B) \to \mathcal{C}/(A\times B)$ is a functor, defined on objects as: $$((X,g),(Y,h)) \mapsto ((X\times_0Y),g\times_1h)$$ and on morphisms as $(f_1,f_2) \mapsto f_1\times_1f_2$. The fact that it is a functor follows from the functoriality of $\times$.

My question is, what are the requirements on $\mathcal{C}$ for $ {}\boxtimes (Z, f) \colon \mathcal{C}/A \to \mathcal{C}/(A\times B)$ to preserve colimits? To avoid making this an XY-question, I want to show that for $\mathcal{C} = \mathsf{Set}$, the partially applied induced product functor distributes over the pointwise coproduct in $\mathsf{Set}/X$. I have proved this manually, but I feel like this should follow from some more general categorical property, and should hold for much more general categories, not just $\mathsf{Set}$. This is the rest of the context:

Let ${+}\colon \mathsf{Set}^2\to \mathsf{Set}$ be the functor that sends objects to their coproduct in $\mathsf{Set}$. Then there is a functor $\oplus\colon (\mathsf{Set}/A)^2 \to \mathsf{Set}/A$ , defined on objects as: $$((X,g),(Y,h)) \mapsto ((X+_0Y),[g,h])$$ and on morphisms as $(f_1,f_2) \mapsto f_1 +_1 f_2$.

The fact that it is a functor follows from the interaction laws of sum and cotupling.

Let $(A,g)\colon \mathsf{Set}/D$, $(B,h_1),(C,h_2)\colon \mathsf{Set}/E$. Then $$ g \boxtimes (h_1 \oplus h_2) \simeq g \boxtimes h_1 \oplus g \boxtimes h_1. $$ ^^This is the theorem I'd like to have a general proof for.

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For any category $\mathcal{C}$ and object $X\in\mathcal{C}$, the forgetful functor $\mathcal{C}_{/X}\rightarrow\mathcal{C}$ both preserves and reflects colimits. In your scenario, there is a commutative diagram $$\require{AMScd} \begin{CD} \mathcal{C}_{/A} @>{-\boxtimes(Z,f)}>> \mathcal{C}_{/A\times B}\\ @VVV @VVV\\ \mathcal{C} @>{-\times Z}>> \mathcal{C}, \end{CD}$$ so the top horizontal map preserves $\mathcal{I}$-shaped colimits if the bottom map preserves $\mathcal{I}$-shaped colimits. In $\mathcal{C}=\mathbf{Set}$, the bottom map, for any set $Z$, preserves all colimits since it is a left-adjoint, i.e. $\mathbf{Set}$ is cartesian-closed.

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