Implication details

Proof: Let $\C$ be a countably extensive category with cofiltered-limit-stable epimorphisms. We show that $\C$ is thin. (With other results, one can further deduce that $\C$ is trivial.)Let $T$ be an object, let $T_m$ be the summand with index $m$ of the coproduct $\coprod_{m \in \IN} T$, and for $n \in \IN$ let $$\textstyle D(n) \coloneqq \coprod_{m \geq n} T_m.$$ The canonical morphisms $D(n') \to D(n)$ for $n \leq n'$ form a cofiltered diagram $D : (\IN,\geq) \to \C$, whose limit is initial by this lemma. For every $n \in \IN$, the codiagonal $D(n) \to T$ is an epimorphism, since its composite with the coproduct inclusion $T_n \hookrightarrow D(n)$ is the identity. Together, they form a morphism from $D$ to the constant diagram with value $T$, whose limit is $T$. By assumption, their limit $0 \to T$ is an epimorphism. Since $0$ is initial, it follows that any two morphisms $T \rightrightarrows Y$ coincide. Thus, $\C$ is thin.