The Stacks project

Proof. Choose a set $\{ L_ i\} _{i \in I}$ of linearly adequate functors such that every linearly adequate functor is isomorphic to one of the $L_ i$. This is possible. Suppose that we can find $Q(F) \to F$ with (1) and (2)' or every $i \in I$ the map $\mathop{\mathrm{Hom}}\nolimits (L_ i, Q(F)) \to \mathop{\mathrm{Hom}}\nolimits (L_ i, F)$ is a bijection. Then (2) holds. Namely, combining Lemmas 46.3.6 and 46.3.11 we see that every adequate functor $G$ sits in an exact sequence

with $K$ and $L$ direct sums of linearly adequate functors. Hence (2)' implies that $\mathop{\mathrm{Hom}}\nolimits (L, Q(F)) \to \mathop{\mathrm{Hom}}\nolimits (L, F)$ and $\mathop{\mathrm{Hom}}\nolimits (K, Q(F)) \to \mathop{\mathrm{Hom}}\nolimits (K, F)$ are bijections, whence the same thing for $G$.

Consider the category $\mathcal{I}$ whose objects are pairs $(i, \varphi )$ where $i \in I$ and $\varphi : L_ i \to F$ is a morphism. A morphism $(i, \varphi ) \to (i', \varphi ')$ is a map $\psi : L_ i \to L_{i'}$ such that $\varphi ' \circ \psi = \varphi$. Set

There is a natural map $Q(F) \to F$, by Lemma 46.3.12 it is adequate, and by construction it has property (2)'. $\square$