Lemma 12.30.2. Let $\mathcal{I}$ be a category. Let $\mathcal{A}$ be an additive, Karoubian category. Let $F : \mathcal{I} \to \mathcal{A}$ and $G : \mathcal{I} \to \mathcal{A}$ be functors. The following are equivalent
$\mathop{\mathrm{colim}}\nolimits _\mathcal {I} F \oplus G$ exists, and
$\mathop{\mathrm{colim}}\nolimits _\mathcal {I} F$ and $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} G$ exist.
In this case $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} F \oplus G = \mathop{\mathrm{colim}}\nolimits _\mathcal {I} F \oplus \mathop{\mathrm{colim}}\nolimits _\mathcal {I} G$.
Proof.
Assume (1) holds. Set $W = \mathop{\mathrm{colim}}\nolimits _\mathcal {I} F \oplus G$. Note that the projection onto $F$ defines natural transformation $F \oplus G \to F \oplus G$ which is idempotent. Hence we obtain an idempotent endomorphism $W \to W$ by Categories, Lemma 4.14.8. Since $\mathcal{A}$ is Karoubian we get a corresponding direct sum decomposition $W = X \oplus Y$, see Lemma 12.4.2. A straightforward argument (omitted) shows that $X = \mathop{\mathrm{colim}}\nolimits _\mathcal {I} F$ and $Y = \mathop{\mathrm{colim}}\nolimits _\mathcal {I} G$. Thus (2) holds. We omit the proof that (2) implies (1).
$\square$
Comments (1)
Comment #541 by Nuno on