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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