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Viewing Version 9 of 'unit of adjunction'
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Title of object: unit of adjunction
Canonical Name: UnitOfAnAdjunction
Type: Definition

Created on: 2007-07-27 21:34:00
Modified on: 2007-07-29 02:32:38

Creator: CWoo
Modifier: CWoo
Author: CWoo

Classification: msc:18A40
Defines: unit, counit
Synonyms: unit of adjunction=co-unit

Preamble:

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Content:

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 $ST(C)$.

\begin{thm} $(T(C),\eta_C)$ is a universal arrow from $C$ to $S$. \end{thm}
\begin{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
$$
\xymatrix{
& C \ar[dr]^{f} \ar[dl]_{\eta_C} & \\
ST(C) \ar[rr]_{S(g)} && S(Y) }$$
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
$$
\xymatrix{
\hom_{\mathcal{D}}(T(C),T(C)) \ar[d]_{\hat{g}} \ar[r]^{\nu_C} & \hom_{\mathcal{C}}(C,ST(C)) \ar[d]^{\overline{g}} \\
\hom_{\mathcal{D}}(T(C),Y) \ar[r]_{\nu_{C,Y}} & \hom_{\mathcal{C}}(C,S(Y)), }$$
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.
\end{proof}
\begin{thm} $\eta: C \mapsto \eta_C$ is a natural transformation from the identity functor $I_{\mathcal{C}}$ to $ST$. \end{thm}
\begin{proof}
Suppose $f:A\to B$ is a morphism in $\mathcal{C}$. We want to show that
$$\xymatrix{
A \ar[d]_{\eta_A} \ar[rr]^f && B \ar[d]^{\mu_B} \\
ST(A) \ar[rr]_{ST(f)} && ST(B) }$$
is commutative. To see this, write out the expressions
\begin{alignat*}{2}
\eta_B\circ f &= \nu_B(1_{T(B)})\circ f & \mbox{definition of }\eta_B \\
&= 1_{ST(B)}\circ \nu_B(1_{T(B)})\circ f & \mbox{property of identity morphism} \\
&= S(1_{T(B)})\circ \nu_B(1_{T(B)})\circ f & \quad \mbox{property of functor on identity morphism} \\
&= \nu_A(1_{T(B)}\circ 1_{T(B)}\circ T(f)) & \mbox{naturality of adjoint functors}\\
&= \nu_A(1_{T(B)}\circ T(f)) & \mbox{idempotency of identity morphism} \\
&= \nu_A(T(f)\circ 1_{T(A)}) & T(f)\mbox{ commutes with identity morphisms}\\
&= \nu_A(T(f)\circ 1_{T(A)}\circ 1_{T(A)}) & \mbox{idempotency of identity morphisms}\\
&= ST(f)\circ \nu_A(1_{T(A)})\circ 1_{T(A)} & \mbox{naturality of adjoint functors}\\
&= ST(f)\circ \nu_A(1_{T(A)}) & \mbox{idempotency of identity morphisms}\\
&= ST(f)\circ \eta_A & \mbox{definition of }\eta_A.
\end{alignat*}
\end{proof}

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

\begin{thebibliography}{9}
\bibitem{Ma}S. Mac Lane, \emph{Categories for the Working Mathematician} (2nd edition), Springer-Verlag, 1997.
\end{thebibliography}