Lemma 4.24.8. 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{\mathrm{Mor}}\nolimits (C, v_1D) \to \mathop{\mathrm{Mor}}\nolimits (C, v_2D) is the map \mathop{\mathrm{Mor}}\nolimits (u_1C, D) \to \mathop{\mathrm{Mor}}\nolimits (u_2C, D) induced by \beta _ C : u_2C \to u_1C. Namely, this means the map
\mathop{\mathrm{Mor}}\nolimits (u_1 v_1 D, D') \to \mathop{\mathrm{Mor}}\nolimits (u_2 v_1 D, D')
induced by \beta _{v_1 D} is the same as the map
\mathop{\mathrm{Mor}}\nolimits (v_1 D, v_1 D') \to \mathop{\mathrm{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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