Lemma 19.14.1. The functor $G$ above has a left adjoint $F : \text{Mod}_ R \to \mathcal{A}$.

**Proof.**
We will give two proofs of this lemma.

The first proof will use the adjoint functor theorem, see Categories, Theorem 4.25.3. Observe that that $G : \mathcal{A} \to \text{Mod}_ R$ is left exact and sends products to products. Hence $G$ commutes with limits. To check the set theoretical condition in the theorem, suppose that $M$ is an object of $\text{Mod}_ R$. Choose a suitably large cardinal $\kappa $ and denote $E$ a set of objects of $\mathcal{A}$ such that every object $A$ with $|A| \leq \kappa $ is isomorphic to an element of $E$. This is possible by Lemma 19.11.4. Set $I = \coprod _{A \in E} \mathop{\mathrm{Hom}}\nolimits _ R(M, G(A))$. We think of an element $i \in I$ as a pair $(A_ i, f_ i)$. Finally, let $A$ be an arbitrary object of $\mathcal{A}$ and $f : M \to G(A)$ arbitrary. We are going to think of elements of $\mathop{\mathrm{Im}}(f) \subset G(A) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(U, A)$ as maps $u : U \to A$. Set

Since $G$ is left exact, we see that $G(A') \subset G(A)$ contains $\mathop{\mathrm{Im}}(f)$ and we get $f' : M \to G(A')$ factoring $f$. On the other hand, the object $A'$ is the quotient of a direct sum of at most $|M|$ copies of $U$. Hence if $\kappa = |\bigoplus _{|M|} U|$, then we see that $(A', f')$ is isomorphic to an element $(A_ i, f_ i)$ of $E$ and we conclude that $f$ factors as $M \xrightarrow {f_ i} G(A_ i) \to G(A)$ as desired.

The second proof will give a construction of $F$ which will show that “$F(M) = M \otimes _ R U$” in some sense. Namely, for any $R$-module $M$ we can choose a resolution

Then we define $F(M)$ by the corresponding exact sequence

This construction is independent of the choice of the resolution and is functorial; we omit the details. For any $A$ in $\mathcal{A}$ we obtain an exact sequence

which is isomorphic to the sequence

which shows that $F$ is the left adjoint to $G$. $\square$

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