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
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
Comments (0)
There are also: