The Stacks project

Proof. We first prove $G$ is fully faithful, or equivalently that $F \circ G \to \text{id}$ is an isomorphism, see Categories, Lemma 4.24.4. First, given an object $A$ the map $F(G(A)) \to A$ is surjective, because every map of $U \to A$ factors through $F(G(A))$ by construction. On the other hand, the map $F(G(A)) \to A$ is the adjoint of the map $\text{id} : G(A) \to G(A)$ and hence injective by Lemma 19.14.2.

The functor $F$ is right exact as it is a left adjoint. Since $\text{Mod}_ R$ has enough projectives, to show that $F$ is exact, it is enough to show that the first left derived functor $L_1F$ is zero. To prove $L_1F(M) = 0$ for some $R$-module $M$ choose an exact sequence $0 \to K \to P \to M \to 0$ of $R$-modules with $P$ free. It suffices to show $F(K) \to F(P)$ is injective. Now we can write this sequence as a filtered colimit of sequences $0 \to K_ i \to P_ i \to M_ i \to 0$ with $P_ i$ a finite free $R$-module: just write $P$ in this manner and set $K_ i = K \cap P_ i$ and $M_ i = \mathop{\mathrm{Im}}(P_ i \to M)$. Because $F$ is a left adjoint it commutes with colimits and because $\mathcal{A}$ is a Grothendieck abelian category, we find that $F(K) \to F(P)$ is injective if each $F(K_ i) \to F(P_ i)$ is injective. Thus it suffices to check $F(K) \to F(P)$ is injective when $K \subset P = R^{\oplus n}$. Thus $F(K) \to U^{\oplus n}$ is injective by an application of Lemma 19.14.2. $\square$