Lemma 14.3.2. Let $\mathcal{C}$ be a category.

1. Given a simplicial object $U$ in $\mathcal{C}$ we obtain a sequence of objects $U_ n = U([n])$ endowed with the morphisms $d^ n_ j = U(\delta ^ n_ j) : U_ n \to U_{n-1}$ and $s^ n_ j = U(\sigma ^ n_ j) : U_ n \to U_{n + 1}$. These morphisms satisfy the opposites of the relations displayed in Lemma 14.2.3, namely

1. If $0 \leq i < j \leq n + 1$, then $d^ n_ i \circ d^{n + 1}_ j = d^ n_{j - 1} \circ d^{n + 1}_ i$.

2. If $0 \leq i < j \leq n - 1$, then $d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_{j - 1} \circ d^{n - 1}_ i$.

3. If $0 \leq j \leq n - 1$, then $\text{id} = d^ n_ j \circ s^{n - 1}_ j = d^ n_{j + 1} \circ s^{n - 1}_ j$.

4. If $0 < j + 1 < i \leq n$, then $d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_ j \circ d^{n - 1}_{i - 1}$.

5. If $0 \leq i \leq j \leq n - 1$, then $s^ n_ i \circ s^{n - 1}_ j = s^ n_{j + 1} \circ s^{n - 1}_ i$.

2. Conversely, given a sequence of objects $U_ n$ and morphisms $d^ n_ j$, $s^ n_ j$ satisfying (1)(a) – (e) there exists a unique simplicial object $U$ in $\mathcal{C}$ such that $U_ n = U([n])$, $d^ n_ j = U(\delta ^ n_ j)$, and $s^ n_ j = U(\sigma ^ n_ j)$.

3. A morphism between simplicial objects $U$ and $U'$ is given by a family of morphisms $U_ n \to U'_ n$ commuting with the morphisms $d^ n_ j$ and $s^ n_ j$.

Proof. This follows from Lemma 14.2.4. $\square$

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