### The conceptual inverse in category theory

Modified: 25 May 2017; Status: completed

Lately I’ve been thinking about induced representations and the Frobenius reciprocity theorem which have a natural interpretation in terms of adjoint functor in category theory. This post is an introduction to the categorical point of view that I need at the level I’m comfortable with. (You can find better general introductions out there by mathematicians such as this 3-parter by Todd Trimble.)

To me, one of the most powerful lessons from category theory is that knowing the relations between an object with other objects is as good as knowing the object itself, as the object can be defined by these relations. A lot of time there is a unique (up to a natural isomorphism which doesn’t entail making any arbitrary choice) object that has a “universal” property that can simulate the relations to all other objects of the same kind and so can be thought of as the “best” object of its kind. As a bonus, all of these can be represented by diagrams of objects and arrows. I will give some examples of these on the way but let us first define what categories are.

A category $\mathcal{C}$ is a class of objects and morphisms between them. If $X$ and $Y$ are objects in $\mathcal{C}$, the class of morphisms between $X$ and $Y$ is denoted by $\text{Hom}(X,Y)$. They are also represented diagrammatically as arrows $f:X \to Y$ for $f \in \text{Hom}(X,Y)$. For $\mathcal{C}$ to be a category, there must be a unit (identity) morphism and the compositions of morphisms must be associative. Here are some categories.

• Set of all sets. Since there is no set of all sets, the word “class” in the definition of a category cannot be replaced by “set”. The morphisms are arbitrary maps between sets.
• Grp of all groups, Ring of all rings, and Field$_p$ of all fields with characteristic $p=0$ or a prime number with the respective homomorphisms as morphisms. (There is no homomorphism between fields of different characteristics.)
• Top of topological spaces with continuous maps as morphisms. In particular, homeomorphisms are isomorphisms.
• Man$^{\infty}$ of smooth manifolds with infinitely differentiable maps as morphisms. In particular, diffeomorphisms are isomorphisms.
• Vec$_k$ of vector spaces over a field $k$ with linear maps as morphisms.
My favorite elementary example of uniquely defining an object by morphisms is the construction of a product and a coproduct. A product of $A$ and $B$ is an object $C$ together with a morphism from $C$ to $A$ and $B$ such that, if $D$ is another object also equipped with some morphisms to $A$ and $B$, then there is a unique morphism from $D$ to $C$ such that the following diagram “commutes” (so we call this kind of diagrams commutative diagrams), meaning that every way to compose such morphisms to go from $D$ to $A$ or $B$ gives the same result.

The rational behind this definition is that $C$ together with the mappings $\alpha$ and $\beta$ is the best object that acts as a product of $A$ and $B$ since giving any morphism from any object $D$ to $A$ or $B$ is equivalent to giving $\alpha$ and $\beta$ from $C$, so we can just forget about $D$ altogether and regard $C$ as a universal simulator of such relations.

The same diagram with all arrows reversed defines a coproduct $C$.

In Set, the product is the direct product $A \times B$ and the coproduct is the disjoint union $A \amalg B$. We will just verify the first statement to familiarize ourselves with the meaning of the diagrams. For the direct product $C=A \times B$, a natural choice for the mappings $\alpha$ and $\beta$ is the projection \begin{aligned} \alpha(a,b) = a, && \beta(a,b) = b. \end{aligned} We have to show that, for any $\alpha'$ and $\beta'$ from a set $D$, there is a morphism $\gamma$ from $D$ to $C$ that makes the diagram commutes and uniquely so. Obviously, for $d\in D$, this map is \begin{aligned} \gamma(d) = (\alpha'(d),\beta'(d)), \end{aligned} If there is another map $\gamma'$ from $D$ to $C$ that makes the diagram commutes, it must sends $d$ to an element of $A \times B$ that projects to $\alpha'(d)$ on $A$ and $\beta'(d)$ on $B$. But that element is none other than $(\alpha'(d),\beta'(d))$. Thus, the uniqueness is established. For the disjoint union $C = A \amalg B$, a natural choice for the mappings $\alpha$ and $\beta$ is the embedding of $A$ and $B$ into the union. Then the proof proceeds in the same manner by making obvious choices and following the arrows around (“diagram chasing”).

We can ask if the direct product and the direct sum of sets are the only product and coproduct objects in Set. The nice answer is that they are. And this is not only true in Set but also in a general category as well. We will not prove that here but it amounts to using the definition of a product or a coproduct and do some more diagram chasing.

Why going so far to define something that should reduce to the notion of a product in any category? Well, this construction guides us when it is not clear what the right structure of the product is. This is the case in the category Top of all topological spaces, which Todd Trimble gives as an example and to which I will not go into since it is new to me so I would just be regurgitating what Todd already said better.

## Functors

Suppose that $\mathcal{C}$ and $\mathcal{D}$ are two categories and $X,Y \in \mathcal{C}$. A (covariant) functor $F$ is a map (of both objects and morphisms) $F: \mathcal{C} \to \mathcal{D}$ such that $\text{Hom}(X,Y) \to \text{Hom}(F(X) , F(Y))$ preserves identity morphisms and compositions.

One of the most trivial kinds of functors are ones that simply throw away information. The forgetful functor Grp $\to$ Set sends a group to its underlying set; it completely forgets the group structure. There are also functors that are partially forgetful such as the functor from Grp to Ab the category of all abelian groups which set $ab=ba$ for all elements $a,b$ in a group.

More interesting is the free functor $F:$ Set $\to$ Grp sending a set to the “free group”. For a set $S$, $F(S)$ is the set of all expressions (“words”) that can be composed from elements of $S$: every power and inverse of each and every element and (noncommutative) products of them; it is the least constrained group that can be built from elements of $S$ as generators. The categorical way to say this is that the free functor has the universal property that any map $\sigma$ from $S$ to some group $G$ factors through the free group $F(S)$ with a unique group homomorphism $\varphi$.

For a given group, the forgetful functor sends it to a particular set, and for a given set, the free functor sends it to a particular group. So they are in some sense inverse functors to each other even though the categories Set and Grp are clearly not the same. This concept of a generalized inverse is formalized in the notion of an adjoint functor.

Functors $F: \mathcal{C} \to \mathcal{D}$ and $G: \mathcal{D} \to \mathcal{C}$ are adjoint functors if for any $X \in \mathcal{C}$ and $Y \in \mathcal{D}$, there is an isomorphism

\begin{aligned} \text{Hom}_{\mathcal{D}} (F(X),Y) \simeq \text{Hom}_{\mathcal{C}} (X,G(Y)). \end{aligned} $F$ is called a left adjoint of $G$ and $G$ is called a right adjoint of $F$.

Why should this be true in the case of the forgetful functor $E$ and the free functor $F$ between Set and Grp? A group homomorphism $\varphi :F(S) \to G$ is completely determined if we know what it does to each element of $S$. In other words, what $\varphi$ really acts on is the underlying set of $G$. That is, giving $\varphi$ is equivalent to giving $\psi : S \to E(G)$. \begin{aligned} \text{Hom}_{\textbf{Grp}} (F(S),G) \simeq \text{Hom}_{\textbf{Set}} (S,E(G)) \end{aligned} so that the free functor is a left adjoint of the forgetful functor.

In representation theory, a restriction of a representation of a group $G$ to a subgroup $H$ is a functor whose left adjoint is the induction of a representation of $H$ to a “free” representation of $G$. There, the Frobenius reciprocity theorem is nothing but the property of adjoint functors. I will talk about how this can be used to deduce the well known fact that each irreducible representation of SO(3) appears only once in the decomposition of functions on a sphere.

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