Lemma 4.16.2. Let $\mathcal{C}$ be a category. Let $X$ be an object of $\mathcal{C}$. Let $M : \mathcal{I} \to \mathcal{C}/X$ be a diagram in the category of objects over $X$. If the index category $\mathcal{I}$ is connected and the limit of $M$ exists in $\mathcal{C}/X$, then the limit of the composition $\mathcal{I} \to \mathcal{C}/X \to \mathcal{C}$ exists and is the same.
Proof. Let $L \to X$ be an object representing the limit in $\mathcal{C}/X$. Consider the functor
\[ W \longmapsto \mathop{\mathrm{lim}}\nolimits _ i \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, M_ i). \]
Let $(\varphi _ i)$ be an element of the set on the right. Since each $M_ i$ comes equipped with a morphism $s_ i : M_ i \to X$ we get morphisms $f_ i = s_ i \circ \varphi _ i : W \to X$. But as $\mathcal{I}$ is connected we see that all $f_ i$ are equal. Since $\mathcal{I}$ is nonempty there is at least one $f_ i$. Hence this common value $W \to X$ defines the structure of an object of $W$ in $\mathcal{C}/X$ and $(\varphi _ i)$ defines an element of $\mathop{\mathrm{lim}}\nolimits _ i \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}/X}(W, M_ i)$. Thus we obtain a unique morphism $\phi : W \to L$ such that $\varphi _ i$ is the composition of $\phi $ with $L \to M_ i$ as desired. $\square$
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