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.
Product in the category of functors.
category-theoryfunctorsnatural-transformations
Related Solutions
I want to point out something potentially misleading about Marek's answer. The n-categories he mentions are not categories, but generalizations of them, so the question still remains, why do categories only form a 2-category, that is, why do people stop after categories, functors, and natural transformations? Why don't people define modifications of natural transformations?
I think it is good to realize that categories really are in an essential way only 2-categorical, if you want interesting higher morphisms you do need to define something like a higher category. One way to think about it is this: natural tranformations are basically homotopies. To make this precise, take I to be the category with two objects, 0 and 1, one morphism from 0 to 1 and the identity morphisms. Then, it is easy to check that to specify a natural tranformation between two functors F and G (both functors C → D) is the same as specifying a functor H : C × I → D which agrees with F on C × {0} and with G on C × {1}.
So then we could get higher morphisms by saying they are homotopies of homotopies, i.e., functors C × I × I with appropriate restrictions. This works, and we indeed get some definition of modification, but it is not interesting as it reduces to just a commuting square of natural transformations, i.e., it can be described simply in terms of the structure we already had.
This is similar to what happens for, say, groups: you can think of a group as a category with a single object where all the morphisms are invertible (the morphisms are the group elements and the composition law is the group product). Then group homomorphisms are simply functors. This makes it sound as if groups now magically have a higher sort of morphisms: natural tranformations between functors! And indeed they do, they are even useful in certain contexts, but they're not terribly interesting: a natural transformation between to group homomorphisms f and g is simply a group element y such that f(x) = y g(x) y -1 . Again, this is described in terms of things we already new about (the group element y and conjugation), and is not really a brand new concept.
A functor from $\mathcal A$ to $\mathcal B$ must take every morphism in $\mathcal A$ to a corresponding morphism in $\mathcal B$.
A natural transformation connects one particular functor $\mathcal A\to\mathcal B$ to one particular other functor $\mathcal A\to\mathcal B$. It does not need to apply to every functor in some category of functors.
Therefore it cannot be thought of as a "functor of functors" -- it does not even take functors as input.
What you can do is think of a natural tranformation as a morphism between functors. All functors $\mathcal A\to\mathcal B$ together with the natural transformations between them make up a functor category.
Best Answer
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.