# How to show two functors form an adjunction

Say I have two functors $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G:\mathcal{D}\rightarrow\mathcal{C}$. How can I show they form an adjunction without writing explicitly the natural transformations $\hom_\mathcal{C}(x,Gy)\cong \hom_\mathcal{D}(Fx,y)$?

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One of the most important and also useful(!) descriptions is via the unit and counit. I will sketch this, details can be found in any book on category theory, or at the nlab.

If $F :C \to D$ and $G : D \to C$ are functors, then $F$ is left adjoint to $G$ iff there are natural transformations $\eta : \mathrm{id}_C \to GF$ (unit) and $\varepsilon : FG \to \mathrm{id}_D$ (counit) such that the triangular identities hold: The compositions $F \xrightarrow{F\eta} FGF \xrightarrow{\varepsilon F} F$ and $G \xrightarrow{\eta G} GFG \xrightarrow{G \varepsilon} G$ equal the identity (on $F$ resp. $G$). There is a nice visualization using string diagrams. Actually this notion applies to every bicategory. Monoidal categories are bicategories with one object, and in that case the above definition gives the notion of dual objects. So in general this definition is some kind of "categorified duality". It has many more advantages in theory, but also in practice.

Here is a typical example (which should be more well-known): Consider the functor $F : \mathsf{Top} \to \mathsf{\mathbb{C}Alg}^{op}$ mapping a topological space $X$ to the $\mathbb{C}$-algebra of continuous functions $C(X,\mathbb{C})$. For a continuous map $f : X \to Y$ we have the pullback homomorphism $F(f) := f^* : C(Y,\mathbb{C}) \to C(X,\mathbb{C})$. Conversely, consider the functor $G : \mathsf{\mathbb{C}Alg}^{op} \to \mathsf{Top}$ which maps a $\mathbb{C}$-algebra $A$ to the set of homomorphisms of $\mathbb{C}$-algebras $\chi : A \to \mathbb{C}$ (characters), endowed with the subspace topology of the product ${\mathbb{C}}^A$. Again the action on morphisms is given by pullback. There is a canonical morphism $\eta_X : X \to G(F(X))$ defined by $\eta_X(x)(f)=f(x)$. There is also a canonical morphism $\epsilon_A : F(G(A)) \to A$, i.e. a homomorphism of $\mathbb{C}$-algebras $A \to F(G(A))$ given by the same formula, i.e. $\epsilon_A(a)(\chi)=\chi(a)$. One checks that the triangular identities are satisfied. Hence, $F$ is left adjoint to $G$.

One of the main purposes of adjunctions is to approximate equivalences of categories. This is made precise by the following easy lemma or exercise:

Lemma. If $F$ is left adjoint to $G$ with unit $\eta$ and counit $\varepsilon$ as above, then $x \in C$ is called a fixed point if $\eta_x : x \to G(F(x))$ is an isomorphism. In the same we define $y \in D$ to be a fixed point if $\varepsilon_y : F(G(y)) \to y$ is an isomorphism. We get full subcategories $\mathrm{Fix}(GF) \subseteq C$ and $\mathrm{Fix}(FG) \subseteq D$, which are preserved by $F$ and $G$. In fact, $F$ and $G$ induce an equivalence of categories $\mathrm{Fix}(GF) \cong \mathrm{Fix}(FG)$.

Applying this to the example above, we get the famous Gelfand duality between compact Hausdorff spaces and commutative unital C*-algebras. Many more equivalences of categories arise this way. I hope that this illustrates the importance of the unit and the counit description.

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+1 nicely described. –  Ehsan M. Kermani Mar 26 '13 at 2:12

In Mac Lane's Categories for the Working Mathematician, on the chapter on adjunctions, there are several equivalent conditions listed for verifying a pair of functors are adjoints. Quite often a very convenient one is to verify the triangular identities, especially when you already have the two functors.

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I happen to have a copy of that book. I'll take a look there, then. –  Daniel Robert-Nicoud Mar 25 '13 at 19:05
@BabyDragon the adjoint functor theorem is a handy tool for showing a functor has a left/right adjoint. But it is not so convenient in verifying that two functors are adjoint. The adjoint functor theorem is a set theoretic construction that gives an adjoint. To use it for OP's purposes, one would need to apply it to $G$, obtain a left adjoint for it in a rather complicated set theoretic formula, and then compare that to the functor $F$. This is likely going to be much more difficult than directly constructing the adjunction between $F$ and $G$. –  Ittay Weiss Mar 25 '13 at 18:57