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.

  • 3
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $F$ 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

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

| cite | improve this answer | |
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ Thanks for that answer from a more abstract perspective! $\endgroup$ – S. M. Roch Sep 21 '19 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.