Lemma 14.19.10. Let \mathcal{C} be a category which has finite limits.
For every n the functor \text{sk}_ n : \text{Simp}(\mathcal{C}) \to \text{Simp}_ n(\mathcal{C}) has a right adjoint \text{cosk}_ n.
For every n' \geq n the functor \text{sk}_ n : \text{Simp}_{n'}(\mathcal{C}) \to \text{Simp}_ n(\mathcal{C}) has a right adjoint, namely \text{sk}_{n'}\text{cosk}_ n.
For every m \geq n \geq 0 and every n-truncated simplicial object U of \mathcal{C} we have \text{cosk}_ m \text{sk}_ m \text{cosk}_ n U = \text{cosk}_ n U.
If U is a simplicial object of \mathcal{C} such that the canonical map U \to \text{cosk}_ n \text{sk}_ nU is an isomorphism for some n \geq 0, then the canonical map U \to \text{cosk}_ m \text{sk}_ mU is an isomorphism for all m \geq n.
Comments (1)
Comment #1023 by correction_bot on
There are also: