Let $\mathcal{C},\mathcal{D}$ be categories and $(T,S,\nu)$ be an adjunction from $\mathcal{C}$ to $\mathcal{D}$ . For every pair of objects $C\in\mathcal{C}$ and $D\in\mathcal{D}$ , we have a bijection \begin{equation} \nu_{C,D}:\hom_{\mathcal{D}}(T(C),D) \longrightarrow \hom_{\mathcal{C}}(C,S(D)) \end{equation}that is natural in each variable.

If we set $D=T(C)$ , and write $\nu_C$ for $\nu_{C,T(C)}$ , then we get a bijection $$\nu_C:\hom_{\mathcal{D}}(T(C),T(C)) \longrightarrow \hom_{\mathcal{C}}(C,ST(C))$$ where $ST$ is the abbreviation of $S\circ T$ .

As $1_{T(C)}$ is the identity morphism in $\hom_{\mathcal{D}}(T(C),T(C))$ , define \begin{equation} \eta_C:=\nu_C(1_{T(C)}). \end{equation}Note that $\eta_C$ is a morphism in $\mathcal{C}$ from $C$ to $ST(C)$ . Also, naturality in $C$ means that if $f:C'\to C$ and $g:T(C)\to T(C')$ , then \begin{equation} Sg\circ \eta_c \circ f=\nu_{C'}(g\circ Tf). \end{equation}

Proof. Let $Y$ be an object in $\mathcal{D}$ and $f:C\to S(Y)$ a morphism in $\mathcal{C}$ . We want to find a morphism $g:T(C)\to Y$ in $\mathcal{D}$ such that

is a commutative diagram. The existence and uniqueness of $g$ is guaranteed by the bijection $$\nu_{C,Y}:\hom_{\mathcal{D}}(T(C),Y) \longrightarrow \hom_{\mathcal{C}}(C,S(Y)),$$ where $f=\nu_{C,Y}(g)$ , and the commutativity of the triangle above is guaranteed by the naturality in the second variable

where $\hat{g}:=\hom_{\mathcal{D}}(1_{T(C)},g)$ and $\overline{g}:=\hom_{\mathcal{C}}(1_C,S(g))$ , as $$\overline{g}\circ \nu_C(1_{T(C)})=\hom_{\mathcal{C}}(C,S(g))\circ \eta_C=S(g)\circ \eta_C$$ on the one hand, and $$\nu_{C,Y}\circ \hat{g}(1_{T(C)})=\nu_{C,Y}\circ \hom(T(C),g)(1_{T(C)})=\nu_{C,Y}(g\circ 1_{T(C)})=\nu_{C,Y}(g)=f$$ on the other, and the two are equal.

Proof. Suppose $f:A\to B$ is a morphism in $\mathcal{C}$ . We want to show that

is commutative. To see this, write out the expressions

		property of identity morphism
		property of functor on identity morphism
		by equation (3) above
		commutes with identity morphisms
		property of functor on identity morphism
		by equation (3) above
		property of identity morphisms

Definition. The natural transformation $\eta:I_{\mathcal{C}}\dot{\to} ST$ defined above is called the unit of the adjunction $(T,S,\nu)$ from $\mathcal{C}$ to $\mathcal{D}$ .

Dually, one can find a natural transformation $\epsilon:TS \dot{\to} I_{\mathcal{D}}$ called the counit of the adjunction $(T,S,\nu):\mathcal{C}\to\mathcal{D}$ . To do this, set $C=S(D)$ and use equation (1) to get a bijection $\nu_D:=\nu_{S(D),D}$ and subsequently define \begin{equation} \epsilon_D:=\nu_D(1_{S(D)}). \end{equation}As in the previous theorems, one can, by reversing all the arrows, show that $(S(D),\epsilon_D)$ is a universal arrow from $D$ to $T$ , and that $\epsilon$ is indeed a natural transformation from $TS$ to $I_{\mathcal{D}}$ .

Bibliography

1

S. Mac Lane, Categories for the Working Mathematician (2nd edition), Springer-Verlag, 1997.