Product in the category of functors.

Question

Let $A$ be a category and $C= Fun(A, Set)$ (i.e. the objects are functors and morphisms are natural transformations between them). I want to know if this category has a product. For given $X \in A$ and functors $F_i \in C$, is $\Pi_{i \in \Lambda}F_i(X)$ the correct product? If yes, what would be the projection maps onto the $F_i$'s? I'm having a hard time constructing the required unique natural transformation. Any help is appreciated.

Answer 1

Yes, the product of $F_i\in\text{Obj}(\text{Fun}(A,\mathbf{Set}))$, $i\in\Lambda$, is the functor $\prod_{i\in\Lambda}F_i$, such that $(\prod_{i\in\Lambda}F_i)(X)=\prod_{i\in\Lambda}(F_i(X))$ for every $X\in\text{Obj}(A)$ and $(\prod_{i\in\Lambda}F_i)(f)=\prod_{i\in\Lambda}(F_i(f))$ for every $f\in\text{Mor}(A)$. Projections are natural transformations $P_j\colon(\prod_{i\in\Lambda}F_i)\to F_j$, $j\in\Lambda$, such that $P_j(X)=p_j\colon(\prod_{i\in\Lambda}(F_i(X)))\to F_j(X)$, $X\in \text{Obj}(A)$, $j\in\Lambda$, where $p_j$ is a projection of an "ordinary" product in $\mathbf{Set}$. These transformations are natural by the universal properties of products (note, that $p_j\circ(\prod_{i\in\Lambda}(F_i(f)))=F_j(f)\circ p_j$, and the morphism $\prod_{i\in\Lambda}(F_i(f))$ is unique with such property). Such products are called pointwise by the obvious reason.

Answer 2

Oskar already gave a good answer (+1). I just wanted to add a different perspective.

Given a collection of objects $C_i$ in a category $C$, the product of the $C_i$, $\prod_i C_i$, is an object representing the (contravariant) functor $C\to \newcommand\Set{\mathbf{Set}}\Set$ $$A \mapsto \prod_i\newcommand\Hom{\operatorname{Hom}} \Hom(A,C_i).$$

Thus if $C$ and $A$ are categories, if $F_i$ are functors in $C^A$, the product of the $F_i$ is the functor $F$ representing $$G \mapsto \prod_i\Hom_{C^A}(G,F_i).$$ I.e. it is a functor satisfying $$\Hom_{C^A}(G,F)\simeq \prod_i \Hom_{C^A}(G,F_i).$$ However if we already know that $C$ has products, then let $F'(x)=\prod_i F_i(x)$ for any $x\in A$.

Then $$\Hom_C(y,F'(x)) \simeq \prod_i \Hom_C(y,F_i(x))$$ as functors from $C^{\mathrm{op}}\times A\to \Set$. Thus for any functor $G:A\to C$, given a natural transformation $\mu:G\to F'$, for any $x\in A$, we can use this natural isomorphism to break apart the component $\mu_x:G(x)\to F'(x)$ of the natural transformation into maps $\mu_{x,i} : G(x)\to F_i(x)$ giving natural transformations $\mu_i$, and conversely given natural transformations $\mu_i : G\to F_i$, we can put them together to get a natural transformation $\mu$. Thus $F'$ satisfies $$\Hom_{C^A}(G,F')\simeq \prod_i\Hom_{C^A}(G,F_i),$$ so $F'=\prod_i F_i$ as desired.

It should hopefully be clear how to generalize this argument to arbitrary codomain categories and arbitrary limits/colimits as indicated by Derek's comment. (Well I already generalized to arbitrary codomain categories, I guess I mean to arbitrary shapes of limits/colimits).