Adjoint Functors

Today we’re going to talk about adjoint functors.

Definition: Let $L:C\to D$ be a functor, and let $R:D\to C$ be another functor. Then $L$ and $R$ are adjoint functors if for $X\in$ Obj $(C)$ and $Y\in$ Ob $(D)$ , we have $Mor_C(X,RY)\simeq Mor_D(LX,Y)$ .

Maclane has famously stated in his seminal book that “adjoint functors arise everywhere”. However, what is the utility of such functors?

1. Solutions to Optimization Problems: Suppose you have an rng (which is a ring without the identity element), and you want to “adjoin” the minimal number of elements to the rng such that it becomes a ring. Explicitly, you want to adjoint $1$ to the rng, along with the elements $r\pm 1$ for all $r\in rng$ . How can you solve this problem? For any fixed ring $R$ , consider the category $E$ in which the objects are morphisms of the form $R\to S_j$ , where $S_j$ are unital rings. A morphism between the objects $R\to S_i$ and $R\to S_j$ is a morphism between $S_i$ and $S_j$ . Then the final object in this category is $R$ adjoined with the unity. This is the smallest ring which contains $R$ . Hence, this is the most efficient solution to our problem.

When we have a functor $L$ , we ask ourselves what is the problem $R$ to which $L$ is the most efficient solution. Then $L$ is the left adjoint of $R$ .

We have different formulations for adjoint functor. The unit-counit formulation says that $1_c=R\circ L$ and $1_D=L\circ R$ .

In terms of Hom isomorphism: An adjunction $L\vdash R$ is a natural isomorphism between functors $C^{op}\times D\to Set$ . What does this mean? We study the isomorphisms between functors from $C^{op}\times D$ to $Set$ . This natural isomorphism is given by $Hom_C(X,RY)\simeq Hom_D(LX,Y)$ . How is this a natural transformation between functors from $C^{op}\times D$ to $Set$ ? So we take a tuple $(X,Y)\in C\times D$ , the make the following two functors act on it: $(X,Y)\to Hom(LX,Y)$ and $(X,Y)\to Hom(X,RY)$ . Then the natural isomorphism is the isomorphism between these two functors. But why $C^{op}$ ? Why not just $C$ ? If one draws the commutative diagram, one will soon figure out that reversing the arrows in $C$ is a simple way of mapping $Hom(RX,Y)$ to $Hom(X,LY)$ .

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. I am always excited about talking to people about their research. Please please set up a meeting with me if you feel that I might have an interesting perspective to offer- https://calendly.com/ayushkhaitan/meeting-with-ayush View more posts

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