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'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: |
2007-07-28 20:17:21 |
| Classification: |
msc:18A40 |
| Defines: |
unit, counit |
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
$$\nu_{C,D}:\hom_{\mathcal{D}}(T(C),D) \longrightarrow \hom_{\mathcal{C}}(C,S(D))$$
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 $$\eta_C:=\nu_C(1_{T(C)}).$$ 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 $\eta_B\circ f=\nu_B(1_{T(B)})\circ f=1_{ST(B)}\circ \nu_B(1_{T(B)})\circ f=S(1_{T(B)})\circ \nu_B(1_{T(B)})\circ f=\nu_A(1_{T(B)}\circ 1_{T(B)}\circ T(f))=\nu_A(1_{T(B)}\circ T(f))=\nu_A(T(f)\circ 1_{T(A)})=\nu_A(T(f)\circ 1_{T(A)}\circ 1_{T(A)})=ST(f)\circ \nu_A(1_{T(A)})\circ 1_{T(A)}=ST(f)\circ \nu_A(1_{T(A)})=ST(f)\circ \eta_A$.
\end{proof}
more to come... |
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