Lemma 19.14.2. Let $f : M \to G(A)$ be an injective map in $\text{Mod}_ R$. Then the adjoint map $f' : F(M) \to A$ is injective too.

Proof. Choose a map $R^{\oplus n} \to M$ and consider the corresponding map $U^{\oplus n} \to F(M)$. Consider a map $v : U \to U^{\oplus n}$ such that the composition $U \to U^{\oplus n} \to F(M) \to A$ is $0$. Then this arrow $v : U \to U^{\oplus n}$ is an element $v$ of $R^{\oplus n}$ mapping to zero in $G(A)$. Since $f$ is injective, we conclude that $v$ maps to zero in $M$ which means that $U \to U^{\oplus n} \to F(M)$ is zero by construction of $F(M)$ in the proof of Lemma 19.14.1. Since $U$ is a generator we conclude that

$\mathop{\mathrm{Ker}}(U^{\oplus n} \to F(M) \to A) = \mathop{\mathrm{Ker}}(U^{\oplus n} \to F(M))$

To finish the proof we choose a surjection $\bigoplus _{i \in I} R \to M$ and we consider the corresponding surjection

$\pi : \bigoplus \nolimits _{i \in I} U \longrightarrow F(M)$

To prove $f'$ is injective it suffices to show that $\mathop{\mathrm{Ker}}(\pi ) = \mathop{\mathrm{Ker}}(f' \circ \pi )$ as subobjects of $\bigoplus _{i \in I} U$. However, now we can write $\bigoplus _{i \in I} U$ as the filtered colimit of its subobjects $\bigoplus _{i \in I'} U$ where $I' \subset I$ ranges over the finite subsets. Since filtered colimits are exact by AB5 for $\mathcal{A}$, we see that

$\mathop{\mathrm{Ker}}(\pi ) = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} \left(\bigoplus \nolimits _{i \in I'} U\right) \bigcap \mathop{\mathrm{Ker}}(\pi )$

and

$\mathop{\mathrm{Ker}}(f' \circ \pi ) = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} \left(\bigoplus \nolimits _{i \in I'} U\right) \bigcap \mathop{\mathrm{Ker}}(f' \circ \pi )$

and we get equality because the same is true for each $I'$ by the first displayed equality above. $\square$

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