Lemma 4.24.7. Let $u_1, u_2 : \mathcal{C} \to \mathcal{D}$ be functors with right adjoints $v_1, v_2 : \mathcal{D} \to \mathcal{C}$. Let $\beta : u_2 \to u_1$ be a transformation of functors. Let $\beta ^\vee : v_1 \to v_2$ be the corresponding transformation of adjoint functors. Then

\[ \xymatrix{ u_2 \circ v_1 \ar[r]_\beta \ar[d]_{\beta ^\vee } & u_1 \circ v_1 \ar[d] \\ u_2 \circ v_2 \ar[r] & \text{id} } \]

is commutative where the unlabeled arrows are the counit transformations.

**Proof.**
This is true because $\beta ^\vee _ D : v_1D \to v_2D$ is the unique morphism such that the induced maps $\mathop{Mor}\nolimits (C, v_1D) \to \mathop{Mor}\nolimits (C, v_2D)$ is the map $\mathop{Mor}\nolimits (u_1C, D) \to \mathop{Mor}\nolimits (u_2C, D)$ induced by $\beta _ C : u_2C \to u_1C$. Namely, this means the map

\[ \mathop{Mor}\nolimits (u_1 v_1 D, D') \to \mathop{Mor}\nolimits (u_2 v_1 D, D') \]

induced by $\beta _{v_1 D}$ is the same as the map

\[ \mathop{Mor}\nolimits (v_1 D, v_1 D') \to \mathop{Mor}\nolimits (v_1 D, v_2 D') \]

induced by $\beta ^\vee _{D'}$. Taking $D' = D$ we find that the counit $u_1 v_1 D \to D$ precomposed by $\beta _{v_1D}$ corresponds to $\beta ^\vee _ D$ under adjunction. This exactly means that the diagram commutes when evaluated on $D$.
$\square$

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