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<record version="13" id="9806">
 <title>unit of adjunction</title>
 <name>UnitOfAnAdjunction</name>
 <created>2007-07-27 21:34:00</created>
 <modified>2007-12-18 11:22:21</modified>
 <type>Definition</type>
<parent id="2691">adjoint functor</parent>
 <creator id="3771" name="CWoo"/>
 <author id="3771" name="CWoo"/>
 <classification>
	<category scheme="msc" code="18A40"/>
 </classification>
 <defines>
	<concept>unit</concept>
	<concept>counit</concept>
 </defines>
 <synonyms>
	<synonym concept="unit of adjunction" alias="co-unit"/>
 </synonyms>
 <related>
	<object name="UniversalProperty"/>
 </related>
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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 $ST(C)$.  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}

\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 
$$
\xymatrix{
&amp; C \ar[dr]^{f} \ar[dl]_{\eta_C} &amp; \\
 ST(C) \ar[rr]_{S(g)} &amp;&amp; S(Y) }$$
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
$$
\xymatrix{
\hom_{\mathcal{D}}(T(C),T(C)) \ar[d]_{\hat{g}} \ar[rr]^{\nu_C} &amp;&amp; \hom_{\mathcal{C}}(C,ST(C)) \ar[d]^{\overline{g}} \\
\hom_{\mathcal{D}}(T(C),Y) \ar[rr]_{\nu_{C,Y}} &amp;&amp; \hom_{\mathcal{C}}(C,S(Y)), }$$
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
$$\xymatrix{
A \ar[d]_{\eta_A} \ar[rr]^f &amp;&amp; B \ar[d]^{\mu_B} \\
ST(A) \ar[rr]_{ST(f)} &amp;&amp; ST(B) }$$
is commutative.  To see this, write out the expressions 
\begin{alignat*}{2}
\eta_B\circ f &amp;= 1_{ST(B)}\circ \eta_B\circ f &amp; \mbox{property of identity morphism} \\ 
&amp;= S(1_{T(B)})\circ \eta_B\circ f &amp; \qquad\qquad\qquad \mbox{property of functor on identity morphism} \\ 
&amp;= \nu_A(1_{T(B)}\circ T(f)) &amp; \mbox{by equation (3) above}\\
&amp;= \nu_A(T(f)\circ 1_{T(A)}) &amp; T(f)\mbox{ commutes with identity morphisms}\\
&amp;= \nu_A(T(f)\circ T(1_A)) &amp; \mbox{property of functor on identity morphism}\\
&amp;= ST(f)\circ \eta_A\circ 1_A &amp; \mbox{by equation (3) above}\\
&amp;= ST(f)\circ \eta_A &amp; \mbox{property of identity morphisms}.
\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 $(T,S,\nu):\mathcal{C}\to\mathcal{D}$.  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 $I_{\mathcal{D}}$.

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