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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