The Stacks project

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

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 $(u, v)$. $\square$