Lemma 4.24.9. Let \mathcal{A}, \mathcal{B}, and \mathcal{C} be categories. Let v : \mathcal{A} \to \mathcal{B} and v' : \mathcal{B} \to \mathcal{C} be functors with left adjoints u and u' respectively. Then
The functor v'' = v' \circ v has a left adjoint equal to u'' = u \circ u'.
Given X in \mathcal{A} we have
4.24.9.1
\begin{equation} \label{categories-equation-compose-counits} \epsilon _ X^ v \circ u(\epsilon ^{v'}_{v(X)}) = \epsilon ^{v''}_ X : u''(v''(X)) \to X \end{equation}
Where \epsilon is the counit of the adjunctions.
Proof.
Let us unwind the formula in (2) because this will also immediately prove (1). First, the counit of the adjunctions for the pairs (u, v) and (u', v') are maps \epsilon _ X^ v : u(v(X)) \to X and \epsilon _ Y^{v'} : u'(v'(Y)) \to Y, see discussion following Definition 4.24.1. With u'' and v'' as in (1) we unwind everything
u''(v''(X)) = u(u'(v'(v(X)))) \xrightarrow {u(\epsilon _{v(X)}^{v'})} u(v(X)) \xrightarrow {\epsilon _ X^ v} X
to get the map on the left hand side of (4.24.9.1). Let us denote this by \epsilon _ X^{v''} for now. To see that this is the counit of an adjoint pair (u'', v'') we have to show that given Z in \mathcal{C} the rule that sends a morphism \beta : Z \to v''(X) to \alpha = \epsilon _ X^{v''} \circ u''(\beta ) : u''(Z) \to X is a bijection on sets of morphisms. This is true because, this is the composition of the rule sending \beta to \epsilon _{v(X)}^{v'} \circ u'(\beta ) which is a bijection by assumption on (u', v') and then sending this to \epsilon _ X^ v \circ u(\epsilon _{v(X)}^{v'} \circ u'(\beta )) which is a bijection by assumption on (u, v).
\square
Comments (2)
Comment #7461 by Dun Liang on
Comment #7612 by Stacks Project on