Lemma 7.5.2. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor between categories. Assume
the category $\mathcal{C}$ has a final object $X$ and $u(X)$ is a final object of $\mathcal{D}$ , and
the category $\mathcal{C}$ has fibre products and $u$ commutes with them.
Then the index categories $(\mathcal{I}^ u_ V)^{opp}$ are filtered (see Categories, Definition 4.19.1).
Proof. The assumptions imply that the assumptions of Lemma 7.5.1 are satisfied (see the discussion in Categories, Section 4.18). By Categories, Lemma 4.19.8 we see that $\mathcal{I}_ V$ is a (possibly empty) disjoint union of directed categories. Hence it suffices to show that $\mathcal{I}_ V$ is connected.
First, we show that $\mathcal{I}_ V$ is nonempty. Namely, let $X$ be the final object of $\mathcal{C}$, which exists by assumption. Let $V \to u(X)$ be the morphism coming from the fact that $u(X)$ is final in $\mathcal{D}$ by assumption. This gives an object of $\mathcal{I}_ V$.
Second, we show that $\mathcal{I}_ V$ is connected. Let $\phi _1 : V \to u(U_1)$ and $\phi _2 : V \to u(U_2)$ be in $\mathop{\mathrm{Ob}}\nolimits (\mathcal{I}_ V)$. By assumption $U_1\times U_2$ exists and $u(U_1\times U_2) = u(U_1)\times u(U_2)$. Consider the morphism $\phi : V \to u(U_1\times U_2)$ corresponding to $(\phi _1, \phi _2)$ by the universal property of products. Clearly the object $\phi : V \to u(U_1\times U_2)$ maps to both $\phi _1 : V \to u(U_1)$ and $\phi _2 : V \to u(U_2)$. $\square$
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