Lemma 14.19.10. Let $\mathcal{C}$ be a category which has finite limits.

1. For every $n$ the functor $\text{sk}_ n : \text{Simp}(\mathcal{C}) \to \text{Simp}_ n(\mathcal{C})$ has a right adjoint $\text{cosk}_ n$.

2. 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 $\text{sk}_{n'}\text{cosk}_ n$.

3. For every $m \geq n \geq 0$ and every $n$-truncated simplicial object $U$ of $\mathcal{C}$ we have $\text{cosk}_ m \text{sk}_ m \text{cosk}_ n U = \text{cosk}_ n U$.

4. 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 $m \geq n$.

Proof. The existence in (1) follows from Lemma 14.19.2 above. Parts (2) and (3) follow from the discussion in Remark 14.19.9. After this (4) is obvious. $\square$

Comment #1023 by correction_bot on

The first sentence of the proof seems strange. Maybe something more like

"The existence in (1) follows from Lemma \ref{lemma-existence-cosk} above. (2) and (3) follow from the discussion in Remark \ref{remark-inductive-coskelet}."

was intended.

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