Coproducts in $\textbf{Set}\times \textbf{Set}$

Question

The answer in this question says that the coproduct $(A_1,B_1)+(A_2,B_2)$ in $\textbf{Set}\times \textbf{Set}$ is $(A_1+A_2,B_1+B_2)$. Why is this the case? I suppose this is a special case of some general result, but I don't understand which one. The closest result about "pointwise computation" that I can think of is that limits/colimits in functor categories are computed pointwise, but there's no functor categories here.

Answer 1

In fact there is. We have an isomorphism of categories $\mathsf{Set} \cong \operatorname{Fun}(*,\mathsf{Set})$. Using this we can deduce $$\begin{align*} \operatorname{Fun}(*+*,\mathsf{Set}) &\cong \operatorname{Fun}(*,\mathsf{Set})\times\operatorname{Fun}(*,\mathsf{Set})\\ &\cong \mathsf{Set}\times\mathsf{Set} \end{align*}$$ So (co)limits in $\mathsf{Set}\times\mathsf{Set}$ are indeed computed pointwise.

Answer 2

You can also argue with product categories directly (no functor categories are required), and in a much more general situation as well. If $(\mathcal{C}_i)_{i \in I}$ is a family of categories with coproducts, then coproducts in $\prod_{i \in I} \mathcal{C}_i$ are computed "pointwise", i.e., for a family $(A_j)_{j \in J}$ in $\prod_{i \in I} \mathcal{C}_i$ we have $$\coprod_{j \in J} A_j = \Bigl(\coprod_{j \in J} A_{ji}\Bigr)_{i \in I},$$ where we abbreviate $A_{ji} := (A_j)_i$. The proof just uses the definitions. Only in $\stackrel{!}{=}$ something is happening, namely that products of sets commute with products of sets. $$\hom\Bigl(\Bigl(\coprod_{j \in J} A_{j,i}\Bigr)_{i \in I},B\Bigr) = \prod_{i \in I} \hom\Bigl(\coprod_{j \in J} A_{ji},B_i\Bigr) = \prod_{i \in I} \prod_{j \in J} \hom(A_{ji},B_i)$$ $$ \stackrel{!}{=} \prod_{j \in J} \prod_{i \in I} \hom(A_{ji},B_i) = \prod_{j \in J} \hom(A_j,B)$$