Lemma 56.2.2. Let $A$, $\mathcal{B}$, $F$ be as in Lemma 56.2.1. Assume $\mathcal{B}$ is additive and $F$ is additive. Then $F'$ is additive and commutes with arbitrary direct sums.

Proof. To show that $F'$ is additive it suffices to show that $F'(M) \oplus F'(M') \to F'(M \oplus M')$ is an isomorphism for any $A$-modules $M$, $M'$, see Homology, Lemma 12.7.1. Write $M = \mathop{\mathrm{colim}}\nolimits _ i M_ i$ and $M' = \mathop{\mathrm{colim}}\nolimits _ j M'_ j$ as filtered colimits of finitely presented $A$-modules $M_ i$. Then $F'(M) = \mathop{\mathrm{colim}}\nolimits _ i F(M_ i)$, $F'(M') = \mathop{\mathrm{colim}}\nolimits _ j F(M'_ j)$, and

\begin{align*} F'(M \oplus M') & = F'(\mathop{\mathrm{colim}}\nolimits _{i, j} M_ i \oplus M'_ j) \\ & = \mathop{\mathrm{colim}}\nolimits _{i, j} F(M_ i \oplus M'_ j) \\ & = \mathop{\mathrm{colim}}\nolimits _{i, j} F(M_ i) \oplus F(M'_ j) \\ & = F'(M) \oplus F'(M') \end{align*}

as desired. To show that $F'$ commutes with direct sums, assume we have $M = \bigoplus _{i \in I} M_ i$. Then $M = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} \bigoplus _{i \in I'} M_ i$ is a filtered colimit. We obtain

\begin{align*} F'(M) & = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} F'(\bigoplus \nolimits _{i \in I'} M_ i) \\ & = \mathop{\mathrm{colim}}\nolimits _{I' \subset I\text{ finite}} \bigoplus \nolimits _{i \in I'} F'(M_ i) \\ & = \bigoplus \nolimits _{i \in I} F'(M_ i) \end{align*}

The second equality holds by the additivity of $F'$ already shown. $\square$

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