Category theory (2022-23): Lecture 2

Functors and examples of functors. The opposite category. Covariant and contravariant functors. Product categories. The hom functor.

Exercise on powerset functors:

I’ll use sets X, Y, Z, and functions f: X -> Y and g: Y -> Z.

1. Call this contravariant functor F1. Define F1(f) thusly: for any subset T of Y (element of PY), F1(f)(T) = f^-1(T) (the pre-image of T under f).

2. Call this functor F2. Define F2 thusly: for any subset S of X (element of PX), F2(f)(S) = f(S).

3. Call this functor F3. Define F3 thusly: for any subset S of X, F3(f)(S) = {elements y of Y such that f^-1({y}) is a subset of S}.

Remarks: If f is an injection, then F3(f) = F2(f); otherwise F3(f)(S) is a proper subset of F2(f)(S) for some subsets S of X.

Of course F1, F2, and F3 are functors (F1 contravariant) because they preserve identity morphisms and composite morphisms. I found that proving composition for F3 was the only part that took a little work.

Solution to the puzzle:

1) Not all endofunctors on Set preserve monomorphisms.

Summary of my proof:

Define F to take each non-empty set X to the singleton set {X} and to take the empty set 0 to the two-element set {0, {0}}. Functions between non-empty sets go to functions between singleton sets. The empty function on 0 goes to the identity function on {0, {0}}. An empty function from 0 to X gets taken to the function from {0, {0}} onto {X}. Those empty functions 0 to X are injections, but F takes them to non-injections.

2) All endofunctors on Set preserve epimorphisms.

Proof:

Any surjecfion f: A -> B corresponds to a function g: B -> A which can be defined by choosing one element of f^-1({b}) and letting that be g(b). Then f°g is the identity function on B, 1_B. So, any endofunctor F on Set must satisfy F(f)°F(g) = F(f°g) = 1_B. This equation can be true only if F(f) is a surjection.