The Stacks project

Lemma 14.20.2. Let $\mathcal{C}$ be a category. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Let $U$ be a simplicial object of $\mathcal{C}$. To give an augmentation of $U$ towards $X$ is the same as giving a morphism $\epsilon _0 : U_0 \to X$ such that $\epsilon _0 \circ d^1_0 = \epsilon _0 \circ d^1_1$.

Proof. Given a morphism $\epsilon : U \to X$ we certainly obtain an $\epsilon _0$ as in the lemma. Conversely, given $\epsilon _0$ as in the lemma, define $\epsilon _ n : U_ n \to X$ by choosing any morphism $\alpha : [0] \to [n]$ and taking $\epsilon _ n = \epsilon _0 \circ U(\alpha )$. Namely, if $\beta : [0] \to [n]$ is another choice, then there exists a morphism $\gamma : [1] \to [n]$ such that $\alpha $ and $\beta $ both factor as $[0] \to [1] \to [n]$. Hence the condition on $\epsilon _0$ shows that $\epsilon _ n$ is well defined. Then it is easy to show that $(\epsilon _ n) : U \to X$ is a morphism of simplicial objects. $\square$


Comments (1)

Comment #1018 by correction_bot on

In the statement of the lemma, should be .


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Your email address will not be published.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 018H. Beware of the difference between the letter 'O' and the digit '0'.