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
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Comment #1018 by correction_bot on