What does it mean for a mapping to be functorial in this given context?

Question

What is meant in the following exercise by functorial on A?

Let $\mathcal{C}$ be a category and $\{B_i\}_{i\in I}\subset\operatorname{Obj}(\mathcal{C})$ so that the category product $\prod_{i\in I} B_i$ with projections $\pi_j: \prod_{i\in I} B_i \to B_j$ for $j\in I$ exists. Prove that for any $A\in\operatorname{Obj}(\mathcal{C})$ there exists a bijection of sets $$\phi: \operatorname{Hom}_{\mathcal{C}}(A, \prod_{i\in I} B_i)\to\prod_{i\in I}\operatorname{Hom}_{\mathcal{C}}(A, B_i)$$which is functorial on $A$.

A natural choice for me would be $\phi: f \mapsto \{\pi_i \circ f\}_{i\in I}$, which is bijective because of the universal property. So what is the property that I need to show now?

As this is a graded homework, please do not post solutions here. I only want to know what functorial means in this given context.

Thanks in advance to all contributors.

Answer 1

As pointed out by Max in the comments, the author means that the isomorphism $\phi$ is natural. If we have a morphism $\alpha : A' \to A$, then the following diagram must commute:

$$\require{AMScd} \begin{CD} \operatorname{Hom}_{\mathcal C}(A, \prod_{i\in I} B_i) @>{\phi}>> \prod_{i\in I} \operatorname{Hom}_{\mathcal C}(A, B_i) \\ @VV{\overline \alpha}V @VV{\overline \alpha}V \\ \operatorname{Hom}_{\mathcal C}(A', \prod_{i\in I} B_i) @>{\phi}>> \prod_{i\in I} \operatorname{Hom}_{\mathcal C}(A', B_i) \end{CD}$$

$\overline \alpha$ is the precomposition with $\alpha$. Similarly, if we have a family of morphisms $\beta_i : B_i \to B'_i$ then the corresponding diagram induced by $\overline \beta_i$ must commute.

Of course, you can view the functor under consideration as a bifunctor and combine the diagrams above into one diagram and prove naturality in one step. The choice is up to you.

Answer 2

Given $\mathcal{C}$, one can construct the category $[\mathcal{C}^\text{opp}, \textbf{Set}]$ of contravariant functors $\mathcal{C} \rightarrow \textbf{Set}$, with as morphisms the natural transformations. Now it turns out this category contains all products. This is so because if $\{F_i\}_{i \in I}$ is a collection of functors we can just use the fact that the product of sets is always defined and define for $X \in \mathcal{C}$ $$(\prod_i F_i) (X)= \prod_i F_i(X)$$, and if $f : Y \rightarrow X$ in $\mathcal{C}$, just define $$\prod_i F_i (f) ((x_i)_{i \in I})=(F_i(f)(x_i))_{i \in I}$$ where $x_i \in F_i(X)$. One can show we also have obvious projection maps $\pi_i \prod_i F_i \rightarrow \pi_i$ and that these together with the product functor do indeed satisfy the universal property of the product. Now of course we already know a a lot of objects in $[\mathcal{C}^\text{opp}, \textbf{Set}]$ , namely the $\text{Hom}(-,X)$ for $X$ in $\mathcal{C}$ an object. The formalisation of what you need to prove is: There is an isomorphism of contravariant functors $$\text{Hom}(-, \prod_i B_i) \cong \prod_i \text{Hom}(-,B_i).$$