Exercises for meeting #3, again in roughly increasing difficulty. 1-5 are recommended.

**Functors Review**

(1)

(a) Show that the covariant power set functor P : Set → Set, which takes sets to their power set (set of all subsets) and whose action on functions is defined by the direct image P(f)(A) = f[A] = { f(x) | x in A } obeys the functor laws: P(id_A) = id_P(A), and P(f;g) = P(f) ; P(g).

(b) Show that the contravariant power set functor P' : Set → Set, which takes sets to their power set and whose action on function is defined by the inverse image P'(f)(A) = f^-1[A] = { x | exists y in A, f(y) = x } obeys the contravariant functor laws: P(id_A) = id_P(A) and P(f;g) = P(g) ; P(f).

(2) Show that the following statements about opposite categories are equivalent:

(a) F is a contravariant functor C → D.

(b) F is a covariant functor C^op → D.

(c) F is a covariant functor C → D^op.

(d) F is a contravariant functor C^op → D^op.

(3) Show that Op : Cat → Cat, which take each category to its opposite category, is a functor (remember that Cat has categories as objects, functors as morphisms). Is this functor covariant or contravariant?

**Natural Transformations**

(4)

(a) Show that there is a natural transformation η : I → P (where I is the identity functor and P is the power set functor defined in exercise 1) which takes a set X to the singleton set {X}.

(b) Show that there is a natural transformation μ which takes a set of sets to their union. (e.g. μ({{1,2},{2,3}}) = {1,2,3} (I am now having you identify the functors between which this is natural)

(5) Finish the details of the Yoneda lemma: Given a category C, an object A of C, and a functor F : C → Set, F(A) ≅ Nat(Hom(A, -), F) (where Nat is the set of natural transformations between the two functors). Note, this is an isomorphism in Set; i.e. a bijection.

**Harder exercises for the enthusiast with copious free time**

(6)

(a) Given functors F,G,H and natural transformations a : F→G and b : G→H, show that there is a natural transformation a;b : F→H. (Hint: to avoid many tedious details, do this by stitching together commutative squares) Show that this notion of composition is associative.

(b) Given categories B,C,D; functors F,G : B→C and F',G' : C→D, and natural transformations a : F→G and b : F'→G', show that there is a natural transformation a•b : F;F' → G;G'.

(c) Show the "interchange law": for natural transformations a,b,c,d with suitable signatures, (a;b)•(c;d) = (a•c);(b•d). (Not so much hard as just an awful lot of information to keep track of. Be diligent.)

(7) Recall the definition of a product category A×B: its objects are pairs of objects (a,b) where a in A and b in B, and its morphisms are pairs of morphisms (f,g) where f in A and g in B, as long as the domains and codomains line up.

(a) Given categories A, B and an object a in A, define the functor (a, –) : B → A×B which takes the object b in B to (a,b) in A×B. Define its action on morphisms and show that it satisfies the functor laws. Clearly the same holds for (–,b).

(b) Given categories A,B,C and functors F,G : A×B → C, show that a transformation t : F → G is natural if and only if it is natural both as a transformation between ((a, –) ; F) → ((a, –) ; G) (which are functors in B → C) for every a in A and as a transformation between ((–,b) ; F) → ((–,b) ; G) (which are functors in A → C) for every b in B.

This is what we mean by a transformation being natural in multiple variables. It is equivalent to consider it natural in each variable individually, fixing the others, or natural in all of them at the same time (using a product category).