Between unrelated pairs of functors and equivalences lies an adjoint pair.

Given $F: \mathcal C \to \mathcal D$ and $G : \mathcal D \to \mathcal C$, we say $F$ is left adjoint to $G$ or, equivalently $G$ is right adjoint to $F$ and write $F \vdash G$ if there exists a natural isomorphism $$\nu_{X,Y} : \operatorname{Hom}_{\mathcal D}(FX,Y) \to \operatorname{Hom}_{\mathcal C}(X,GY).$$

## Examples

• $\operatorname{Id} \vdash \operatorname{Id}$
• $F \vdash F^{-1}$: let $\nu : F^{-1}F \to \operatorname{Id}$ denote the natural isomorphism. Then

gives the required natural isomorphism $\mu$.

• Let $$\mathfrak f: \operatorname{Grp} \to \operatorname{Set}$$ be the forgetful functor, which takes a group to its underlying set. We also the the free group construction $$F : \operatorname{Set} \to \operatorname{Grp}$$ Then, $\mathfrak f \vdash F$.
• Another forgetful functor forgets the $R$-module structure $$\mathfrak f: \operatorname{Mod-}R \to \operatorname{Set}$$ whose adjoint is the free $R$-module construction $$R^\bullet : \operatorname{Set} \to \operatorname{Mod-}R$$ The natural isomorphism is $$\operatorname{Hom}_R(R^X, M) \to \operatorname{Hom}_{\operatorname{Set}}(X,M) \\ \varphi \mapsto \left( x \mapsto \varphi(e_x) \right)$$ where $e_x, x \in X$ is the $R$-basis for $R^X$.

## Units and counits

Given an adjunction $F \vdash G$, we can canonical maps $$\eta_X := \nu(1_{FX}) : X \to GFX$$ giving a natural transformation $\eta : \operatorname{Id} \to GF$ called the unit of the adjunction.

Similarly, $$\epsilon_Y := \nu^{-1}(1_GY) : FGY \to Y$$ gives the counit $\epsilon : FG \to \operatorname{Id}$ of the adjunction.

The diagrams

and

both commute.

Theorem $F \vdash G$ if and only if there are natural transformations $\eta : \operatorname{Id} \to GF$ and $\epsilon: FG \to \operatorname{Id}$ making the diagrams above commute.

Recall that given a right $R$-module $M$ and a left $R$-module $N$ we can form the tensor product $$M \otimes_R N := \mathbb{Z}^{M \times N}/I$$ where $I$ is the ideal generated by $(mr,n)-(m,rn)$, $(m_1+m_2,n)-(m_1,n)-(m_2,n)$ and $(m,n_1+n_2)-(m,n_1)-(m,n_2)$.

Given a $R$-$S$-bimodule $P$, we have a map $$\nu_{N,M}: \operatorname{Hom}_R(P \otimes_S N, M) \to \operatorname{Hom}_S(N, \operatorname{Hom}_R(P,M)) \\ \phi \mapsto \left( n \mapsto \left(p \mapsto \phi(p \otimes n) \right)\right)$$ Let’s check this givens an adjunction $P \otimes_S - \vdash \operatorname{Hom}_R(P,-)$.

First, because $\phi$ is a $R$-module homomorphism, so is $p \mapsto \phi(p \otimes n)$ for any $n \in N$. So the function is well-defined.

Next, for $n \to (p \mapsto \phi(p \otimes n))$ to be an $S$-module homomorphism we need $$\phi(ps \otimes n) = \phi(p \otimes sn)$$ which is guaranteed by the construction of the tensor product.

Now let’s check this is a bijection. Assume we have $\psi : N \to \operatorname{Hom}_R(P,M)$. Define $$\phi(p \otimes n) := \psi(n)(p)$$ and $$\phi\left(\sum_i p_i \otimes n_i \right) := \sum \phi(p_i \otimes n_i).$$ We leave it as an exercise to show that $\psi \mapsto \phi$ is well-defined and inverse to $\nu$.

## Three functors from a ring morphism

Assume we have a ring homomorphism $f : R \to S$. This equips $S$ with the structure of $S$-$R$ bimodule $$(s,r) \cdot s^\prime = s s^\prime f(r)$$ We get an adjunction $S \otimes_R - \vdash \operatorname{Hom}_S(S,-)$. We already saw that $\operatorname{Hom}_S(S,-) \cong \operatorname{Id}$. Restricting the $R$-action via $f$, we have $$\operatorname{Hom}_S(S,-) \cong f_\ast$$ We will write $$f^\ast M := S \otimes_R M$$ and call it pullback via $f$.

The pushforward $f_\ast$ is also isomorphic to $S \otimes_S -$ (where we have reversed the actions now). Thus, we get another adjunction $$f_\ast \vdash \operatorname{Hom}_R(S,-)$$ We will write $f^! M := \operatorname{Hom}_R(S,M)$ and call it the twisted pullback.

Thus, from $f: R \to S$, we get a chain of adjunctions $$f^\ast \vdash f_\ast \vdash f^!.$$