6.1.5 Composition of Adjunctions
We now show that the formation of right and left adjoints is compatible with composition of $1$-morphisms.
Construction 6.1.5.1. Let $\operatorname{\mathcal{C}}$ be a $2$-category containing objects $C$, $D$, and $E$, together with $1$-morphisms
\[ f: C \rightarrow D \quad \quad f': D \rightarrow E \quad \quad g: D \rightarrow C \quad \quad g': E \rightarrow D \]
and $2$-morphisms $\eta : \operatorname{id}_{C} \Rightarrow g \circ f$ and $\eta ': \operatorname{id}_{D} \Rightarrow g' \circ f'$. We let $c(\eta , \eta ')$ denote the $2$-morphism given by the composition
\[ \operatorname{id}_{C} \xRightarrow {\eta } g \circ f \xRightarrow {\sim } g \circ (\operatorname{id}_{D} \circ f) \xRightarrow {\eta '} g \circ ((g' \circ f') \circ f) \xRightarrow {\sim } g \circ (g' \circ (f' \circ f)) \xRightarrow {\sim } (g \circ g') \circ (f' \circ f), \]
where the unlabeled isomorphisms are given by the unit and associativity constraints of $\operatorname{\mathcal{C}}$. We will refer to $c(\eta , \eta ')$ as the contraction of $\eta $ and $\eta '$.
Proposition 6.1.5.3. Let $\operatorname{\mathcal{C}}$ be a $2$-category containing objects $C$, $D$, and $E$, together with $1$-morphisms
\[ f: C \rightarrow D \quad \quad g: D \rightarrow C \quad \quad f': D \rightarrow E \quad \quad g': E \rightarrow D \]
and $2$-morphisms $\eta : \operatorname{id}_{C} \Rightarrow g \circ f$ and $\eta ': \operatorname{id}_{D} \Rightarrow g' \circ f'$. Let $T$ be another object of $\operatorname{\mathcal{C}}$ equipped with $1$-morphisms $c: T \rightarrow C$ and $e: T \rightarrow E$. Then the diagram
6.1
\begin{equation} \begin{gathered}\label{equation:contraction-compose-adjuncts} \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,E)}( (f' \circ f) \circ c, e ) \ar [dd] \ar [r]_{\sim }^{\alpha _{f',f,c}} & \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,E)}( f' \circ (f \circ c), e ) \ar [d] \\ & \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D)}( f \circ c, g' \circ e ) \ar [d] \\ \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( c, (g \circ g') \circ e ) \ar [r]^-{\alpha _{g,g',e}}_{\sim } & \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( c, g \circ (g' \circ e) ) } \end{gathered} \end{equation}
is commutative. Here the right vertical maps are given by the formation of right adjuncts with respect to $\eta $ and $\eta '$ (in the sense of Construction 6.1.2.1), while the left vertical map is given by the formation of right adjuncts with respect to the contraction $c(\eta ,\eta ')$ of Construction 6.1.5.1.
Proof.
Fix a $2$-morphism $\beta : (f'\circ f) \circ c \Rightarrow e$ in $\operatorname{\mathcal{C}}$. Clockwise and counterclockwise composition around the outside of the diagram (6.1) determines two elements of $\operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( c, (g \circ g') \circ e)$, and we wish to prove that these two elements are the same. Unwinding the definitions, we see that these elements can be obtained as the vertical composition of $c \xRightarrow [\sim ]{\lambda _{c}^{-1}} \operatorname{id}_{C} \circ c \xRightarrow {\eta \circ \operatorname{id}_ c} (g\circ f) \circ c$ with $2$-morphisms given by clockwise and counterclockwise composition around the outside of the diagram
\[ \xymatrix@R =50pt@C=50pt{ (gf)c \ar@ {=>}[d]_-{ \lambda _{f}^{-1} }^{\sim } \ar@ {=>}[r]^{\sim } & g (f c) \ar@ {=>}[d]_-{\lambda _{f}^{-1}}^{\sim } \ar@ {=>}[dr]^-{ \lambda _{f c}^{-1}}_{\sim } & & \\ (g (\operatorname{id}_ D f)) c \ar@ {=>}[d]^-{\eta '} \ar@ {=>}[r]^{\sim } & g ((\operatorname{id}_ D f) c) \ar@ {=>}[r]^{\sim } \ar@ {=>}[d]^-{\eta '} & g (\operatorname{id}_ D (f c)) \ar@ {=>}[d]^-{\eta '} & \\ (g ((g' f') f)) c \ar@ {=>}[d]^-{\sim } \ar@ {=>}[r]^{\sim } & g (((g' f') f) c) \ar@ {=>}[d]^-{\sim } \ar@ {=>}[r]^{\sim } & g ((g' f') (f c)) \ar@ {=>}[r]^{\sim } & g (g' (f' (f c))) \ar@ {=>}[dl]^{\sim } \\ (g(g'(f'f)))c \ar@ {=>}[d]^-{\sim } \ar@ {=>}[r]^{\sim } & g ((g' (f' f)) c) \ar@ {=>}[r]^{\sim } & g (g' ((f' f) c) ) \ar@ {=>}[d]^-{\sim } \ar@ {=>}[r]^{\beta } & g (g' e) \ar@ {=>}[d] \\ ( (g g') (f' f) ) c \ar@ {=>}[rr]^{\sim } & & (g g') (( f' f) c) \ar@ {=>}[r]^{\beta } & (g g') e} \]
in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T, C)$; here denote the composition of $1$-morphisms $u$ and $v$ in $\operatorname{\mathcal{C}}$ by $uv$ (rather than $u \circ v$) to simplify the notation, and the unlabeled isomorphisms are given by the associativity constraints of $\operatorname{\mathcal{C}}$. It will therefore suffice to observe that this diagram is commutative. The commutativity of the pentagonal regions follows from the pentagon identity in $\operatorname{\mathcal{C}}$, the commutativity of the triangle from Proposition 2.2.1.16, and the commutativity of each square from the naturality of the associativity constraints of $\operatorname{\mathcal{C}}$.
$\square$
Corollary 6.1.5.4. Let $\operatorname{\mathcal{C}}$ be a $2$-category containing objects $C$, $D$, and $E$, together with $1$-morphisms
\[ f: C \rightarrow D \quad \quad g: D \rightarrow C \quad \quad f': D \rightarrow E \quad \quad g': E \rightarrow D \]
and $2$-morphisms $\eta : \operatorname{id}_{C} \Rightarrow g \circ f$ and $\eta ': \operatorname{id}_{D} \Rightarrow g' \circ f'$. If $\eta $ and $\eta '$ are units of adjunctions, then the contraction $c(\eta ,\eta '): \operatorname{id}_{C} \Rightarrow (g \circ g') \circ (f' \circ f)$ is also the unit of an adjunction.
Proof.
Combine Proposition 6.1.5.3 with the criterion of Proposition 6.1.2.9.
$\square$
From Corollary 6.1.5.4, we can extract the following slightly less precise consequence:
Corollary 6.1.5.5. Let $\operatorname{\mathcal{C}}$ be a $2$-category containing objects $C$, $D$, and $E$, together with $1$-morphisms
\[ f: C \rightarrow D \quad \quad g: D \rightarrow C \quad \quad f': D \rightarrow E \quad \quad g': E \rightarrow D. \]
If $f$ is left adjoint to $g$ and $f'$ is left adjoint to $g'$, then $f' \circ f$ is left adjoint to $g \circ g'$.
Corollary 6.1.5.6. Let $\operatorname{\mathcal{C}}$ be a $2$-category containing $1$-morphisms $u: C \rightarrow D$ and $v: D \rightarrow E$. If $u$ and $v$ admit left adjoints, then $v \circ u$ admits a left adjoint. If $u$ and $v$ admit right adjoints, then $v \circ u$ admits a right adjoint.
We can also formulate a more precise version of Corollary 6.1.5.4, which explicitly describes the counit of a composite adjunction. For this, we need a variant of Construction 6.1.5.1:
Construction 6.1.5.7. Let $\operatorname{\mathcal{C}}$ be a $2$-category containing objects $C$, $D$, and $E$, together with $1$-morphisms
\[ f: C \rightarrow D \quad \quad g: D \rightarrow C \quad \quad f': D \rightarrow E \quad \quad g': E \rightarrow D \]
and $2$-morphisms $\epsilon : f \circ g \Rightarrow \operatorname{id}_{D}$ and $\epsilon ': f' \circ g' \Rightarrow \operatorname{id}_{E}$ be $2$-morphisms of $\operatorname{\mathcal{C}}$. We let $c( \epsilon , \epsilon ')$ denote the $2$-morphism given by the composition
\[ (f' \circ f) \circ (g \circ g') \xRightarrow {\sim } f' \circ (f \circ (g \circ g')) \xRightarrow {\sim } f' \circ ((f \circ g) \circ g') \xRightarrow {\epsilon } f' \circ (\operatorname{id}_{D} \circ g') \xRightarrow {\sim } f' \circ g' \xRightarrow {\epsilon '} \operatorname{id}_{E} \]
We will refer to $c(\epsilon ,\epsilon ')$ as the contraction of $\epsilon $ and $\epsilon '$.
Corollary 6.1.5.9. Let $\operatorname{\mathcal{C}}$ be a $2$-category containing objects $C$, $D$, and $E$, together with $1$-morphisms
\[ f: C \rightarrow D \quad \quad g: D \rightarrow C \quad \quad f': D \rightarrow E \quad \quad g': E \rightarrow D. \]
Let $(\eta ,\epsilon )$ be an adjunction between $f$ and $g$, and let $(\eta ',\epsilon ')$ be an adjunction between $f'$ and $g'$. Then the pair $( c(\eta , \eta '), c(\epsilon , \epsilon ') )$ is an adjunction between $f' \circ f$ and $g \circ g'$. Here $c(\eta ,\eta ')$ is the contraction of $\eta $ with $\eta '$ (in the sense of Construction 6.1.5.1), and $c(\epsilon , \epsilon ')$ is the contraction of $\epsilon $ with $\epsilon '$ (in the sense of Construction 6.1.5.7).
Proof.
By virtue of Proposition 6.1.5.3 and Corollary 6.1.2.6, it will suffice to show that for every object $T \in \operatorname{\mathcal{C}}$ equipped with $1$-morphisms $c: T \rightarrow C$ and $e: T \rightarrow E$, the diagram
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,E)}( (f' \circ f) \circ c, e ) \ar [r]_{\sim }^{\alpha _{f',f,c}} & \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,E)}( f' \circ (f \circ c), e ) \\ & \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D)}( f \circ c, g' \circ e ) \ar [u] \\ \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( c, (g \circ g') \circ e ) \ar [r]^-{\alpha _{g,g',e}}_{\sim } \ar [uu] & \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( c, g \circ (g' \circ e) ) \ar [u] } \]
commutes, where the right vertical maps are given by the formation of left adjuncts with respect to $\epsilon $ and $\epsilon '$, and the left vertical map is given by the formation of left adjuncts with respect to the contraction $c(\epsilon ,\epsilon ')$ of Construction 6.1.5.7. This follows by applying Proposition 6.1.5.3 to the conjugate $2$-category $\operatorname{\mathcal{C}}^{\operatorname{c}}$.
$\square$