Tail coproducts in extensive categories

Proof. Recall that the initial object $0$ is strict (see here), i.e. every morphism $A \to 0$ is an isomorphism. Since coproducts are disjoint, coproduct inclusions are monomorphisms, so that we may treat them as inclusions of subobjects. In particular, $D(\alpha)$ is a subobject of $D(0)$, which is a coproduct of $D(\alpha)$ and $\coprod_{\beta < \alpha} T_\beta$. Moreover, the limit $L$ of $D$ is the intersection $\bigcap_{\alpha < \kappa} D(\alpha)$ of subobjects of $D(0)$.

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. $\square$