# Kerodon

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Let $\operatorname{\mathcal{C}}$ be a $2$-category and let $f: C \rightarrow D$ be an isomorphism in $\operatorname{\mathcal{C}}$, so that $f$ admits a homotopy inverse $g: D \rightarrow C$ (Definition 2.2.8.17). Then the $1$-morphism $g$ is both right adjoint and left adjoint to $f$. More precisely, we have the following:

Proposition 6.1.4.1. Let $\operatorname{\mathcal{C}}$ be a $2$-category, let $f: C \rightarrow D$ and $g: D \rightarrow C$ be $1$-morphisms of $\operatorname{\mathcal{C}}$, and let $\eta : \operatorname{id}_{C} \Rightarrow g \circ f$ be a $2$-morphism of $\operatorname{\mathcal{C}}$. Assume that either $f$ or $g$ is an isomorphism in $\operatorname{\mathcal{C}}$. Then $\eta$ is the unit of an adjunction (in the sense of Definition 6.1.2.10) if and only if it is an isomorphism in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(C,C)$.

We will give the proof of Proposition 6.1.4.1 at the end of this section.

Corollary 6.1.4.2. Let $\operatorname{\mathcal{C}}$ be a $2$-category and let $f: C \rightarrow D$ be an isomorphism in $\operatorname{\mathcal{C}}$. Then any homotopy inverse to $f$ is both a left adjoint and a right adjoint of $f$.

Proof. Let $g: D \rightarrow C$ be a homotopy inverse to $f$, so that there exists an isomorphism $\eta : \operatorname{id}_{C} \xRightarrow {\sim } g \circ f$ in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(C,C)$. It follows from Proposition 6.1.4.1 that $\eta$ is the unit of an adjunction, and therefore exhibits $g$ as a right adjoint to $f$. A similar argument shows that $g$ is left adjoint to $f$. $\square$

Remark 6.1.4.3. Let $\operatorname{\mathcal{C}}$ be a $2$-category and let $f: C \rightarrow D$ be a $1$-morphism of $\operatorname{\mathcal{C}}$. By definition, $f$ is an isomorphism if and only if there exists a $1$-morphism $g: D \rightarrow C$ together with isomorphisms

$\eta : \operatorname{id}_{C} \xRightarrow {\sim } g \circ f \quad \quad \epsilon : f \circ g \xRightarrow {\sim } \operatorname{id}_{D}$

in the categories $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(C,C)$ and $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(D,D)$, respectively. The main content of Proposition 6.1.4.1 is that, if such isomorphisms exist, then we can choose $\eta$ and $\epsilon$ to be compatible in the sense that they satisfy conditions $(Z1)$ and $(Z2)$ of Definition 6.1.1.1. Note that in this case, $\eta$ is determined by $\epsilon$ and vice versa (Proposition 6.1.2.9).

Corollary 6.1.4.4. Let $\operatorname{\mathcal{C}}$ be a $2$-category, let $f: C \rightarrow D$ and $g: D \rightarrow C$ be $1$-morphisms of $\operatorname{\mathcal{C}}$, and let $(\eta , \epsilon )$ be an adjunction between $f$ and $g$. The following conditions are equivalent:

$(1)$

The $1$-morphism $f$ is an isomorphism in $\operatorname{\mathcal{C}}$.

$(2)$

The $1$-morphism $g$ is an isomorphism in $\operatorname{\mathcal{C}}$.

$(3)$

The $2$-morphisms $\eta$ and $\epsilon$ are isomorphisms in $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(C,C)$ and $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(D,D)$, respectively. In particular, $f$ and $g$ are homotopy inverse to one another.

Proof. The implication $(3) \Rightarrow (1)$ and $(3) \Rightarrow (2)$ are immediate from the definitions, and the reverse implications follow by applying Proposition 6.1.4.1 to $\operatorname{\mathcal{C}}$ and the conjugate $2$-category $\operatorname{\mathcal{C}}^{\operatorname{c}}$. $\square$

Warning 6.1.4.5. In the situation of Corollary 6.1.4.4, it is possible for the unit $\eta : \operatorname{id}_{C} \Rightarrow g \circ f$ to be an isomorphism while the counit $\epsilon : f \circ g \Rightarrow \operatorname{id}_{D}$ is not, or vice versa (in which case, the $1$-morphisms $f$ and $g$ cannot be isomorphisms).

We will deduce Proposition 6.1.4.1 from the following more general result:

Proposition 6.1.4.6. Let $\operatorname{\mathcal{C}}$ be a $2$-category, let $f: C \rightarrow D$ and $g: D \rightarrow C$ be $1$-morphisms of $\operatorname{\mathcal{C}}$, and let $\eta : \operatorname{id}_{\operatorname{\mathcal{C}}} \Rightarrow g \circ f$ be a $2$-morphism of $\operatorname{\mathcal{C}}$ which satisfies the following conditions:

• The $2$-morphisms

$(\operatorname{id}_{f} \circ \eta ): f \circ \operatorname{id}_{C} \Rightarrow f \circ (g \circ f) \quad \quad (\eta \circ \operatorname{id}_{g}): \operatorname{id}_{C} \circ g \Rightarrow (g \circ f) \circ g$

are isomorphisms.

• For every object $T \in \operatorname{\mathcal{C}}$, the composition functor $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( T, D) \xrightarrow { g \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)$ is fully faithful.

Then $\eta$ is the unit of an adjunction $(\eta , \epsilon )$. Moreover, the counit map $\epsilon : f \circ g \Rightarrow \operatorname{id}_{D}$ is an isomorphism.

Proof. Since postcomposition with $g$ induces a fully faithful functor

$\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( D, D ) \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( D, C ),$

there is a unique $2$-morphism $\epsilon : f \circ g \Rightarrow \operatorname{id}_{D}$ for which the horizontal composition $\operatorname{id}_{g} \circ \epsilon$ is equal to the composite map

$g \circ (f \circ g) \xRightarrow {\alpha _{g,f,g} } (g \circ f) \circ g \xRightarrow { ( \eta \circ \operatorname{id}_ g)^{-1} } \operatorname{id}_{C} \circ g \xRightarrow { \lambda _{g} } g \xRightarrow { \rho _{g}^{-1} } g \otimes \operatorname{id}_{D}.$

Moreover, $\epsilon$ is an isomorphism and the pair $(\eta , \epsilon )$ automatically satisfies condition $(Z2)$ of Definition 6.1.1.1. Let $\beta$ denote the composition

$f \xRightarrow [\sim ]{\rho _{f}^{-1}} f \circ \operatorname{id}_{C} \xRightarrow {\operatorname{id}_ f \circ \eta } f \circ (g \circ f) \xRightarrow [\sim ]{\alpha _{f,g,f}} (f \circ g) \circ f \xRightarrow {\epsilon \circ \operatorname{id}_ f} \operatorname{id}_{D} \circ f \xRightarrow [\sim ]{\lambda _{f}} f$

Since $\epsilon$ and $\operatorname{id}_{f} \circ \eta$ are isomorphisms, it follows that $\beta$ is an isomorphism. Applying Corollary 6.1.2.8, we see that $\beta ^2 = \beta$, so that $\beta = \operatorname{id}_ f$. $\square$

Proof of Proposition 6.1.4.1. Let $\operatorname{\mathcal{C}}$ be a $2$-category, let $f: C \rightarrow D$ and $g: D \rightarrow C$ be $1$-morphisms in $\operatorname{\mathcal{C}}$, and assume that $g$ is an isomorphism (the case where $f$ is an isomorphism can be treated by applying a similar argument in the opposite $2$-category $\operatorname{\mathcal{C}}^{\operatorname{op}}$). Suppose first that $\eta : \operatorname{id}_{C} \Rightarrow g \circ f$ is an isomorphism in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(C,C)$. It follows that the horizontal compositions

$(\operatorname{id}_{f} \circ \eta ): f \Rightarrow f \circ (g \circ f) \quad \quad (\eta \circ \operatorname{id}_{g}): g \Rightarrow (g \circ f) \circ g$

are isomorphisms in $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(C,D)$ and $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(D,C)$, respectively. For each object $T \in \operatorname{\mathcal{C}}$, our assumption that $g$ is an isomorphism guarantees that the composition functor $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( T, D) \xrightarrow { g \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)$ is an equivalence of categories, and therefore fully faithful. Invoking the criterion of Proposition 6.1.4.7, we conclude that $\eta$ is the unit of an adjunction.

We now prove the converse. Suppose that the $2$-morphism $\eta : \operatorname{id}_{C} \Rightarrow g \circ f$ is the unit of an adjunction. Our assumption that $g$ is an isomorphism guarantees that we can choose a $1$-morphism $f': C \rightarrow D$ and an isomorphism $\eta ': \operatorname{id}_{C} \xRightarrow {\sim } g \circ f'$ in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(C,C)$. It follows from the first part of the proof that $\eta '$ is the unit of an adjunction. Applying Corollary 6.1.3.7, we deduce that there is a unique isomorphism $\beta : f \xRightarrow {\sim } f'$ for which $\eta '$ is equal to the composition

$\operatorname{id}_{C} \xRightarrow {\eta } g \circ f \xRightarrow {\operatorname{id}_ g \circ \beta } g \circ f'.$

Since $\eta '$ and $(\operatorname{id}_ g \circ \beta )$ are isomorphisms in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(C,C)$, it follows that $\eta$ is also an isomorphism. $\square$

We close this section by proving a converse of Proposition 6.1.4.6, which characterizes adjunctions $(\eta , \epsilon )$ for which the counit $\epsilon$ is an isomorphism.

Proposition 6.1.4.7. Let $\operatorname{\mathcal{C}}$ be a $2$-category, let $f: C \rightarrow D$ and $g: D \rightarrow C$ be $1$-morphisms of $\operatorname{\mathcal{C}}$, and let $(\eta , \epsilon )$ be an adjunction between $f$ and $g$. The following conditions are equivalent:

$(1)$

The $2$-morphism $\epsilon : f \circ g \Rightarrow \operatorname{id}_ D$ is an isomorphism.

$(1')$

The $1$-morphism $f \circ g$ is an isomorphism.

$(2)$

The $2$-morphisms

$(\operatorname{id}_{f} \circ \eta ): f \circ \operatorname{id}_{C} \Rightarrow f \circ (g \circ f) \quad \quad (\eta \circ \operatorname{id}_{g}): \operatorname{id}_{C} \circ g \Rightarrow (g \circ f) \circ g$

are isomorphisms. Moreover, for every object $T \in \operatorname{\mathcal{C}}$, the functor $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C) \xrightarrow { f \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D)$ is essentially surjective.

$(2')$

The $2$-morphism $\eta \circ \operatorname{id}_{g}: \operatorname{id}_{C} \circ g \Rightarrow (g \circ f) \circ g$ is an isomorphism. Moreover, for every object $T \in \operatorname{\mathcal{C}}$, the composite functor

$\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T, D) \xrightarrow { g \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( T, C) \xrightarrow { f \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( T, D )$

is essentially surjective.

$(3)$

For every object $T \in \operatorname{\mathcal{C}}$, the functor $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( T, D) \xrightarrow { g \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)$ is fully faithful.

$(3')$

The functor $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(D,D) \xrightarrow { g \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(D,C)$ is fully faithful.

Proof. We first show that $(1)$ and $(1')$ are equivalent. If $\epsilon$ is an isomorphism, then $f \circ g$ is isomorphic to $\operatorname{id}_{D}$ (as an object of the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}(D,D)$) and is therefore an isomorphism of $\operatorname{\mathcal{C}}$ (Remark 2.2.8.23). Conversely, suppose that $f \circ g$ is an isomorphism. Then it is invertible when viewed as an object of the monoidal category $\underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)$. Since $(\eta , \epsilon )$ is an adjunction, we can regard $f \circ g$ as a coalgebra object of $\underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)$ with counit $\epsilon$ (see Remark ). Applying Proposition 2.1.5.23 (to the monoidal category $\underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)^{\operatorname{op}}$), we deduce that $\epsilon$ is an isomorphism.

We now show that $(1)$ implies $(2)$. Assume that $\epsilon : f \circ g \Rightarrow \operatorname{id}_{D}$ is an isomorphism. Axiom $(Z1)$ of Definition 6.1.1.1 guarantees that the composition

$f \xRightarrow {\sim } f \circ \operatorname{id}_{C} \xRightarrow {\operatorname{id}_ f \circ \eta } f \circ (g \circ f) \xRightarrow {\sim } (f \circ g) \circ f \xRightarrow [\sim ]{\epsilon \circ \operatorname{id}_ f} \operatorname{id}_{D} \circ f \xRightarrow {\sim } f$

is equal to the identity $2$-morphism $\operatorname{id}_{f}$, which proves that the horizontal composition $\operatorname{id}_{f} \circ \eta$ is an isomorphism in $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(C,D)$. Similarly, it follows from axiom $(Z2)$ of Definition 6.1.1.1 that the horizontal composition $\eta \circ \operatorname{id}_{g}$ is an isomorphism in $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(D,C)$. For every $1$-morphism $d: T \rightarrow D$ in $\operatorname{\mathcal{C}}$, the map

$f \circ (g \circ d) \xRightarrow [\sim ]{ \alpha _{f,g,d} } (f \circ g) \circ d \xRightarrow [\sim ]{\epsilon \circ \operatorname{id}_ d} \operatorname{id}_ D \circ d \xRightarrow [\sim ]{ \lambda _ d } d$

is an isomorphism, so that $d$ belongs to the essential image of the functor $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C) \xrightarrow { f \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D)$.

We now show that $(2)$ implies $(2')$. Let $d: T \rightarrow D$ be a $1$-morphism of $\operatorname{\mathcal{C}}$. If the functor $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C) \xrightarrow { f \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D)$ is essentially surjective, then $d$ is isomorphic to $f \circ c$ for some $1$-morphism $c: T \rightarrow C$ of $\operatorname{\mathcal{C}}$. If $\operatorname{id}_{f} \circ \eta$ is an isomorphism, then the chain of isomorphisms

$f \circ c \xRightarrow {\sim } (f \circ \operatorname{id}_{C}) \circ c \xRightarrow {(\operatorname{id}_ f \circ \eta ) \circ \operatorname{id}_ c} (f \circ (g \circ f) ) \circ c \xRightarrow {\sim } f \circ ( (g \circ f) \circ c) \xRightarrow {\sim } f \circ (g \circ (f \circ c) )$

shows that $d$ belongs to the essential image of the composite functor

$\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T, D) \xrightarrow { g \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( T, C) \xrightarrow { f \circ } \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( T, D ).$

We next show that $(2')$ implies $(3)$. Fix an object $T \in \operatorname{\mathcal{C}}$ and a pair of $1$-morphisms $d,d': T \rightarrow D$; we wish to show that the composition map

$\operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D) }( d', d) \rightarrow \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C) }( g \circ d', g \circ d)$

is a bijection. By virtue of assumption $(2')$, we may assume that $d' = f \circ c$, where $c: T \rightarrow C$ is a $1$-morphism of the form $g \circ d''$. By virtue of Proposition 6.1.2.9, the composition

\begin{eqnarray*} \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D)}( f \circ c, d ) & \rightarrow & \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( g \circ (f \circ c), g \circ d ) \\ & \simeq & \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( (g \circ f) \circ c, g \circ d ) \\ & \xrightarrow { \eta \circ \operatorname{id}_ c } & \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( \operatorname{id}_ C \circ c, g \circ d ) \\ & \simeq & \operatorname{Hom}_{\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( c, g \circ d ) \end{eqnarray*}

is a bijection. It will therefore suffice to show that the $2$-morphism $(\eta \circ \operatorname{id}_ c): \operatorname{id}_{C} \circ c \Rightarrow (g \circ f) \circ c$ is an isomorphism. This follows from assumption $(2')$, since $(\eta \circ \operatorname{id}_ c)$ can be rewritten as a composition

$\operatorname{id}_{C} \circ (g \circ d'') \xRightarrow {\sim } (\operatorname{id}_ C \circ g) \circ d'' \xRightarrow {(\eta \circ \operatorname{id}_ g) \circ \operatorname{id}_{d''} } ((g \circ f) \circ g) \circ d'' \simeq (g \circ f) \circ (g \circ d'').$

The implication $(3) \Rightarrow (3')$ is clear. We will complete the proof by showing that $(3')$ implies $(1)$. Assume that $(3')$ is satisfied; we wish to show that the $2$-morphism $\epsilon : f \circ g \Rightarrow \operatorname{id}_{D}$ is an isomorphism. To prove this, it will suffice to show that for every $1$-morphism $u: D \rightarrow D$, vertical precomposition with $\epsilon$ induces a bijection $\operatorname{Hom}_{ \underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)}( \operatorname{id}_ D, u ) \rightarrow \operatorname{Hom}_{ \underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)}( f \circ g, u )$. We now observe that this map fits into a commutative diagram

$\xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{ \underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)}( \operatorname{id}_ D, u ) \ar [r]^-{\epsilon } \ar [d]^{\operatorname{id}_ g \circ } & \operatorname{Hom}_{ \underline{\operatorname{End}}_{\operatorname{\mathcal{C}}}(D)}( f \circ g, u ) \ar [d] \\ \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(D,C)}( g \circ \operatorname{id}_ D, g \circ u ) \ar [r]^-{\sim } & \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(D,C)}( g, g \circ u ) }$

where the bottom horizontal map is induced by the right unit constraint $\rho _{g}: g \circ \operatorname{id}_ D \xRightarrow {\sim } g$, the right vertical map is given by the formation of right adjuncts with respect to $\eta$ (and is therefore bijective by virtue of Corollary 6.1.2.6), and the left vertical map is bijective by virtue of assumption $(3')$. $\square$