The Stacks project

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. $\square$


Comments (1)

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.

There are also:

  • 4 comment(s) on Section 14.19: Coskeleton functors

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 018B. Beware of the difference between the letter 'O' and the digit '0'.