Corollary 6.111.
Assume that \(T\) is both bounded and coherent. Then \(T\) is locally coherent.
Proof
Since \(T\) is bounded, it is generated under colimits by truncated objects. Hence every object of \(T\) admits an effective epimorphism from a coproduct of truncated objects, and it is enough to cover a truncated object \(X \in T_{\leq n}\) by coherent objects. Since \(T\) is coherent, we may find an effective epimorphism \(\bigsqcup_i U_i \twoheadrightarrow X\) where each \(U_i\) is \((n+1)\)-coherent. Because \(X\) is \(n\)-truncated, this map factors through an effective epimorphism \(\bigsqcup_i \tau_n(U_i) \twoheadrightarrow X\). By Proposition 6.108(1) and (3), each \(\tau_n(U_i)\) is coherent.