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