Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Corollary 6.1.2.6. 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 suppose we are given $2$-morphisms $\eta : \operatorname{id}_{C} \Rightarrow g \circ f$ and $\epsilon : f \circ g \Rightarrow \operatorname{id}_{D}$. The following conditions are equivalent:

$(1)$

The pair $(\eta , \epsilon )$ is an adjunction between $f$ and $g$ (in the sense of Definition 6.1.1.1).

$(2)$

For every object $T \in \operatorname{\mathcal{C}}$ and every pair of $1$-morphisms $c: T \rightarrow C$ and $d: T \rightarrow D$, the formation of left and right adjuncts (Construction 6.1.2.1) supplies mutually inverse bijections

\[ \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,D)}( f \circ c, d ) \simeq \operatorname{Hom}_{ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(T,C)}( c, g \circ d ). \]

Proof. The implication $(1) \Rightarrow (2)$ follows from Proposition 6.1.2.5. For the converse, we first observe that $\eta : \operatorname{id}_{C} \Rightarrow g \circ f$ is equal to the right adjunct of the right unit constraint $\rho _{f}: f \circ \operatorname{id}_{D} \xRightarrow {\sim } f$ with respect to $\eta $ (Example 6.1.2.2). If assumption $(2)$ is satisfied, then $\rho _{f}$ is the left adjunct of $\eta $ with respect to $\epsilon $. Similarly, assumption $(2)$ guarantees that $\rho _{g}^{-1}: g \xRightarrow {\sim } g \circ \operatorname{id}_{D}$ is the right adjunct of $\epsilon $ with respect to $\eta $, so that the pair $(\eta , \epsilon )$ is an adjunction by virtue of Example 6.1.2.3. $\square$