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?

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

| cite | improve this answer | |
  • $\begingroup$ So it's essentially: $c = c_F = \{c_x \in C | x \in X, \text{with the transformation $F(x) = c_x$}\}$? $\endgroup$ – Oliver G Feb 2 at 16:07
  • $\begingroup$ 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). $\endgroup$ – Captain Lama Feb 2 at 16:14
  • $\begingroup$ 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. $\endgroup$ – Oliver G Feb 2 at 16:17
  • $\begingroup$ You should rather say "indexed by $X$", but yes. $\endgroup$ – Captain Lama Feb 2 at 16:31

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.