Implication details

Proof: Let $\C$ be an infinitary extensive category which is $\kappa$-coaccessible for a regular cardinal $\kappa$. Recall that the initial object $0$ is strict (see here), i.e. every morphism $A \to 0$ is an isomorphism. We claim that every $\kappa$-copresentable object $X$ is terminal. Since every object is a limit of $\kappa$-copresentable objects, and a limit of terminal objects is terminal, it follows that every object is terminal. In particular, $0$ is terminal, so that every object $A$ admits a morphism $A \to 0$, and hence $A \cong 0$ by strictness. Thus, $\C$ is trivial.To prove the claim, let $X$ be $\kappa$-copresentable and let $T$ be an object. We show that there is exactly one morphism $T \to X$. Let $T_\beta$ be the summand with index $\beta$ of the coproduct $\coprod_{\beta < \kappa} T$. Since coproducts are disjoint, coproduct inclusions are monomorphisms, so that we may treat them as inclusions of subobjects. In particular, for $\alpha < \kappa$ the tail coproduct $$\textstyle D(\alpha) \coloneqq \coprod_{\alpha \leq \beta < \kappa} T_\beta$$ is a subobject of $D(0) = \coprod_{\beta < \kappa} T$, which is a coproduct of $D(\alpha)$ and $\coprod_{\beta < \alpha} T_\beta$. Since $\kappa$ is regular, $(\kappa,\leq)$ is $\kappa$-filtered, so its dual $(\kappa,\geq)$ is $\kappa$-cofiltered. Hence, the inclusions $D(\alpha) \subseteq D(\alpha')$ for $\alpha \geq \alpha'$ form a $\kappa$-cofiltered diagram $D : (\kappa,\geq) \to \C$. Its limit $L$ exists since $\C$ has $\kappa$-cofiltered limits, and it is the intersection $\bigcap_{\alpha < \kappa} D(\alpha)$ of subobjects of $D(0)$.We show that $L \cong 0$ with the same argument as in $\Set$. Since coproducts are stable under pullbacks, we have $L \cong \coprod_{\beta < \kappa} (L \cap T_\beta)$. Since a coproduct of initial objects is initial, it suffices to show that $L \cap T_\beta \cong 0$ for every $\beta < \kappa$. Since $D(0)$ is a coproduct of $D(\beta + 1)$ and $\coprod_{\gamma \leq \beta} T_\gamma$, and coproducts are disjoint, the intersection of these two subobjects is $0$. Since $L \subseteq D(\beta + 1)$ and $T_\beta \subseteq \coprod_{\gamma \leq \beta} T_\gamma$, it follows that $L \cap T_\beta$ admits a morphism to $0$, so that $L \cap T_\beta \cong 0$ by strictness.Since $X$ is $\kappa$-copresentable, the canonical map $$\textstyle \colim_\alpha \Hom(D(\alpha),X) \to \Hom(0,X) = \{\ast\}$$ is bijective. By the universal property of the coproduct, we have $\Hom(D(\alpha),X) \cong \Hom(T,X)^{[\alpha,\kappa[}$, and the transition maps correspond to the restriction maps. Hence, $$\textstyle \colim_\alpha \Hom(T,X)^{[\alpha,\kappa[} \cong \{\ast\}.$$ Since $\Hom(T,X)$ embeds into each $\Hom(T,X)^{[\alpha,\kappa[}$ via constant functions, compatibly with the restriction maps, and restrictions of distinct constant functions remain distinct, $\Hom(T,X)$ embeds into the colimit. Hence, there is at most one morphism $T \to X$. If there were none, then each $\Hom(T,X)^{[\alpha,\kappa[}$ and hence the colimit would be empty. Thus, there is exactly one morphism $T \to X$.