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 $\alpha :  \to [n]$ and taking $\epsilon _ n = \epsilon _0 \circ U(\alpha )$. Namely, if $\beta :  \to [n]$ is another choice, then there exists a morphism $\gamma :  \to [n]$ such that $\alpha$ and $\beta$ both factor as $ \to  \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. $\square$

Comment #1018 by correction_bot on

In the statement of the lemma, $x \in \Ob(\mathcal{C})$ should be $X \in \Ob(\mathcal{C})$.

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).