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

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.

• Authors who say "functorial" often mean "natural". That means that the family $(\phi_A)_A$ is a natural transformation between the obvious functors – Maxime Ramzi Sep 20 '19 at 21:38
• Good to know! But for me it's not obvious: When you say a natural transformation, between which functors that go between which categories? – S. M. Roch Sep 20 '19 at 22:07

## 2 Answers

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.

• Just to note: the question is just about "functorial in $A$". So for the homework OP only has to consider the case with $\bar{\alpha}$ you mentioned, and not the $\beta_i$ case (although it might be a nice exercise to do that too). – Mark Kamsma Sep 21 '19 at 0:04
• Right, thanks for the note! – Ayman Hourieh Sep 21 '19 at 12:48
• Thanks a lot. However, it is still not clear to me why that means that $\phi$ (or anything else) is a natural transformation of a functor $F: \mathcal{C}_1\to\mathcal{C}_2$ to a functor $G: \mathcal{C}_1\to\mathcal{C}_2$. What are $F,G,\mathcal{C}_1, \mathcal{C}_2$? – S. M. Roch Sep 21 '19 at 16:18
• $F$ is a functor from $\mathcal C$ to $\mathrm{Set}$ mapping an object $A$ to $\operatorname{Hom}_{\mathcal C}(A, \prod_{i\in I} B_i)$. Likewise, $G$ maps $A$ to $\prod_{i\in I} \operatorname{Hom}_{\mathcal C}(A, B_i)$. – Ayman Hourieh Sep 21 '19 at 16:31
• That makes it clear to me now. Thanks a lot for your effort! – S. M. Roch Sep 21 '19 at 17:37

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 $$\text{Hom}(-, \prod_i B_i) \cong \prod_i \text{Hom}(-,B_i).$$

• The OP explicitly asked for solutions not to be posted. They just wanted someone to clarify what functorial meant in the exercise. – Ayman Hourieh Sep 20 '19 at 22:31
• This is not a solution – M. Van Sep 20 '19 at 22:33
• Thanks for that answer from a more abstract perspective! – S. M. Roch Sep 21 '19 at 17:38