The Stacks project

Proof. Assume (1)(a), i.e., $\mathcal{I}$ is filtered and $M$ is essentially constant. Let $X = \mathop{\mathrm{colim}}\nolimits M_ i$. Choose $i$ and $X \to M_ i$ as in Categories, Definition 4.22.1. Let $\mathcal{I}'$ be the full subcategory consisting of objects which are the target of a morphism with source $i$. Suppose $i' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I}')$ and choose a morphism $i \to i'$. Then $X \to M_ i \to M_{i'}$ composed with $M_{i'} \to X$ is the identity on $X$. As $\mathcal{A}$ is Karoubian, we find a direct summand decomposition $M_{i'} = X_{i'} \oplus Z_{i'}$, where $Z_{i'} = \mathop{\mathrm{Ker}}(M_{i'} \to X)$ and $X_{i'}$ maps isomorphically to $X$. Pick $i \to k$ and $i' \to k$ such that $M_{i'} \to X \to M_ i \to M_ k$ equals $M_{i'} \to M_ k$ as in Categories, Definition 4.22.1. Then we see that $M_{i'} \to M_ k$ annihilates $Z_{i'}$. Thus (1)(b) holds.

Assume (1)(b), i.e., $\mathcal{I}$ is filtered and we have $\mathcal{I}' \subset \mathcal{I}$ and for $i' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I}')$ a direct sum decomposition $M_{i'} = X_{i'} \oplus Z_{i'}$ as stated in the lemma. To see that $M$ is essentially constant we can replace $\mathcal{I}$ by $\mathcal{I}'$, see Categories, Lemma 4.22.11. Pick any $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and denote $X \to M_ i$ the inverse of the isomorphism $X_ i \to X$ followed by the inclusion map $X_ i \to M_ i$. If $j$ is a second object, then choose $j \to k$ such that $Z_ j \to M_ k$ is zero. Since $\mathcal{I}$ is filtered we may also assume there is a morphism $i \to k$ (after possibly increasing $k$). Then $M_ j \to X \to M_ i \to M_ k$ and $M_ j \to M_ k$ both annihilate $Z_ j$. Thus after postcomposing by a morphism $M_ k \to M_ l$ which annihilates the summand $Z_ k$, we find that $M_ j \to X \to M_ i \to M_ l$ and $M_ j \to M_ l$ are equal, i.e., $M$ is essentially constant.

The proof of (2) is dual. $\square$