# Understanding the functor category $\text{C}^X$ for $X$ a discrete set

From Mac Lane's Category Theory:

Let $$C$$ be a category and let $$X$$ be a discrete set and let $$A = \text{C}^X$$ be the functor category.

Then $$A$$ has as its objects the $$X$$-indexed families $$a = \{a_x : x \in X\}$$ of objects of $$C$$.

I'm having a difficult time seeing how the objects of $$A$$ are $$x$$-indexed families of $$C$$.

If a functor $$(F : X \rightarrow \text{C})$$ is an object of $$A$$, then it's defined entirely on where it sends each $$x \in X$$. So I can see how an object of $$A$$ should be a collection of objects of $$C$$ that are defined by the particular functor $$F$$, i.e., an object of $$A$$ should be $$a_F = \{c \in C : F(x) = c\ \text{for some x \in X}\}$$.

But I'm not seeing how the objects are $$x$$ indexed families.

Can someone explain what I'm missing in the analogy here?

• I guess you mean $C=\operatorname{Set}$? – Captain Lama Feb 2 at 15:53
• I mistyped that, $C$ should just be a category, not one in particular. – Oliver G Feb 2 at 15:54

You are confusing the functor $$F$$ and its image in $$C$$.
Just like a function $$f:X\to Y$$ between sets is not the set $$\{y\in Y\,|\, f(x)=y \text{ for some } x\in X\}$$, a functor $$F:X\to C$$ is not $$\{c\in C\,|\, F(x)=c \text{ for some } x\in X\}$$.
Rather, $$F:X\to C$$ is the association of some $$c\in C$$ for each $$x\in X$$, so we may write $$c_x\in C$$ for this particular element, and this gives a family $$\{c_x\in C\,|\, x\in X\}$$. The fact that $$X$$ is discrete as a category implies that the functor is determined by just this family (there is no need to worry about the action on morphisms).
• So it's essentially: $c = c_F = \{c_x \in C | x \in X, \text{with the transformation$F(x) = c_x$}\}$? – Oliver G Feb 2 at 16:07
• I don't understand what you mean by "transformation". When $X$ is discrete, a functor $F:X\to C$ is the same thing as a simple function from the set $X$ to the set of objects of $C$. A function is the same thing as a family (by definition). – Captain Lama Feb 2 at 16:14
• I will be more clear. What I mean is: each object is a functor, and each set indexed by $x$ depends on the functor chosen. – Oliver G Feb 2 at 16:17
• You should rather say "indexed by $X$", but yes. – Captain Lama Feb 2 at 16:31