MAT 4162 - Homework Assignment 2

Instructions: Pick at least 5 problems of varying length and diﬃculty. You do not have to provide full details in all of the problems, but be sure to indicate which details you declare trivial.

Due date: Monday June 22nd, 4pm.

The category of relations Let Rel be the category whose objects are sets and whose morphisms X → Y are relations R ⊆ X × Y . Two relations R ⊆ X × Y and S ⊆ Y × Z may be composed via S ◦ R = {(x, z)|∃y ∈ Y.R(x, y) ∧ S(y, z). Exercise 1. Show that this composition is associative, and that Rel is indeed a category.

The powerset functor(s) Consider the powerset functor P : Set → Set. On objects, it sends a set X to its powerset P(X). A function f : X → Y is sent to

P(f): P(X) → P(Y ); U 7→ P(f)(U) = f[U] =def {f(x)|x ∈ U}. Exercise 2. Show that P is indeed a functor.

For every set X we have a singleton map ηX : X → P(X), deﬁned by 2 S x 7→ {x}. Also, we have a union map µX : P (X) → P(X), deﬁned by α 7→ α.

Exercise 3. Show that η and µ constitute natural transformations 1Set → P and P2 → P, respectively. Objects of the form P(X) are more than mere sets. In class you have seen that they are in fact complete boolean algebras. For now, we regard P(X) as a complete sup-lattice. A partial ordering (P, ≤) is a complete sup-lattice if it is equipped with a supremum map W : P(P ) → P which sends a subset U ⊆ P to W P , the least upper bound of U in P . Exercise 4. Show that P(X) is a complete sup-lattice. Also show that if (P, ≤, W) is a complete sup-lattice, then the ordering may be recovered from the map W. A morphism of complete sup-lattices (P, W) → (Q, W) is a function f : P → Q such that _ _ f( U) = {f(x)|x ∈ U}.

1 Exercise 5. Show that a morphism of complete sup-lattices automatically pre- serves the ordering. Show that complete sup-lattices and their morphisms form a category, called Sup. We shall now set up an adjunction between the category of sup-lattices and the category of sets. In one direction there is a forgetful functor U : Sup → Set, sending a sup-lattice (P, W) to the underlying set P . In the other direction we have the powerset functor, which sends a set X to the complete sup-lattice P(X). Exercise 6. Prove that the powerset functor is left adjoint to the forgetful functor. The above result explains what the universal property of the powerset oper- ation is: it transforms a set into the free complete sup-lattice on that set! We now connect this with the category of relations Rel. Exercise 7. Show that a relation R ⊆ X × Y is the same thing as a function X → P(Y ) by setting up a bijection between the homsets Rel(X,Y ) and Set(X, P(Y )). Show that this bijection is natural in X and Y . Exercise 8. Show that any relation R ⊆ X × Y gives rise to a function P(R): P(X) → P(Y ), deﬁned by [ P(R)(U) = r(U), x∈U where r : X → P(Y ) is the function which corresponds to R under the bijection from the previous exercise. Show also that this makes P into a functor Rel → Set. The previous exercises prove that there is an adjunction between the category of sets and the category of relations, with left adjoint the inclusion I : Set → Rel, which is the identity on objects and which sends a function to its graph, and with right adjoint the powerset functor. Thus both Rel and mathsfSup have an adjunction with Set, and one won- ders how these two categories relate. Exercise 9. Show that the functor from Rel to Sup which takes X to P(X) (what does this do on arrows?) makes the following diagram commute: Rel / Sup D DD yy DD yy P DD yy U D! |yy Set Show that the functor Rel → Sup is full and faithful, and that its image consists of the free sup-lattices. We wrap up by saying that the category of relations is equivalent to the category of free complete sup-lattices.

2 Other problems Exercise 10. Show that there is a natural bijection between the collection of subsets of X and the homset Rel(1,X) (where 1 is a singleton set). Conclude that the functor Rel → Sets from above is representable.

Exercise 11. Consider Grph, the category of directed graphs and graph ho- momorphisms. By forgetting identities and composition we have a forgetful functor U : Cat → Grph. This functor has a left adjoint which takes a graph G to the free category on G. The objects of this category are the vertices of G, and an arrow f : X → Y is a path in G which starts at X and ends at Y . Prove that this indeed gives a category, and that the construction gives a left adjoint to the forgetful functor. Exercise 12. Let X and Y be sets, and consider the powersets P(X), P(Y ). If f : X → Y is a function we may deﬁne a function

f ∗ : P(Y ) → P(X); f ∗(U) = f −1(U) = {x ∈ X|f(x) ∈ U}.

Note that f ∗ goes in the other direction! Show that this makes P into a functor Set → Setop. The functor from the previous exercise is called the contravariant powerset functor.

Exercise 13. Let again X,Y be sets, and regard now P(X) and P(Y ) as posets, hence as categories. For f : X → Y a function show that f ∗ (which is now regarded as functor) has a left adjoint. (Hint: consider the function P(f).)