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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.

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

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    $\begingroup$ 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). $\endgroup$ – Mark Kamsma Sep 21 '19 at 0:04
  • $\begingroup$ Right, thanks for the note! $\endgroup$ – Ayman Hourieh Sep 21 '19 at 12:48
  • $\begingroup$ 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 $F,G,\mathcal{C}_1, \mathcal{C}_2$? $\endgroup$ – S. M. Roch Sep 21 '19 at 16:18
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    $\begingroup$ $F$ 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 $\prod_{i\in I} \operatorname{Hom}_{\mathcal C}(A, B_i)$. $\endgroup$ – Ayman Hourieh Sep 21 '19 at 16:31
  • $\begingroup$ That makes it clear to me now. Thanks a lot for your effort! $\endgroup$ – S. M. Roch Sep 21 '19 at 17:37
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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 $$\text{Hom}(-, \prod_i B_i) \cong \prod_i \text{Hom}(-,B_i).$$

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

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