To see that we obtain a simplicial object we have to check that the relations (1)(a) -- (e) of Lemma \ref{lemma-characterize-simplicial-object} are satisfied. We will use the short hand notation $$ 1_a = 1_{X_{a - 1}} = 1_Y \star \ldots \star 1_Y \quad (a\text{ factors}) $$ for $a \geq 0$. With this notation we have $$ d^n_j = 1_j \star d \star 1_{n - j} \quad\text{and}\quad s^n_j = 1_j \star s \star 1_{n - j} $$ We are repeatedly going to use the rule that for transformations of funtors $a, a', b, b'$ we have $(a' \circ a) \star (b' \circ b) = (a' \star b') \circ (a \star b)$ provided that the $\star$ and $\circ$ compositions in this formula make sense, see Categories, Lemma \ref{categories-lemma-properties-2-cat-cats}. \medskip\noindent Condition (1)(a) always holds (no conditions needed on $d$ and $s$). Namely, let $0 \leq i < j \leq n + 1$. We have to show that $d^n_i \circ d^{n + 1}_j = d^n_{j - 1} \circ d^{n + 1}_i$, i.e., $$ (1_i \star d \star 1_{n - i}) \circ (1_j \star d \star 1_{n + 1 - j}) = (1_{j - 1} \star d \star 1_{n + 1 - j}) \circ (1_i \star d \star 1_{n + 1 - i}) $$ We can rewrite the left hand side as \begin{align*} & (1_i \star d \star 1_{j - i - 1} \star 1_{n + 1 - j}) \circ (1_i \star 1_1 \star 1_{j - i - 1} \star d \star 1_{n + 1 - j}) \\ & = 1_i \star \left((d \star 1_{j - i - 1}) \circ (1_1 \star 1_{j - i - 1} \star d)\right) \star 1_{n + 1 - j} \\ & = 1_i \star d \star 1_{j - i - 1} \star d \star 1_{n + 1 - j}  \end{align*} The second equality is true because $d \circ 1_1 = d$ and $1_{j - i} \circ (1_{j - i - 1} \star d) = 1_{j - i - 1} \star d$. A similar computation gives the same result for the right hand side. \medskip\noindent We check condition (1)(b). Let $0 \leq i < j \leq n - 1$. We have to show that $d^n_i \circ s^{n - 1}_j = s^{n - 2}_{j - 1} \circ d^{n - 1}_i$, i.e., $$ (1_i \star d \star 1_{n - i}) \circ (1_j \star s \star 1_{n - 1 - j}) = (1_{j - 1} \star s \star 1_{n - 1 - j}) \circ (1_i \star d \star 1_{n - 1 - i}) $$ By the same kind of calculus as in case (1)(a) both sides simplify to $1_i \star d \star 1_{j - i - 1} \star s \star 1_{n - j - 1}$. \medskip\noindent We check condition (1)(c). Let $0 \leq j \leq n - 1$. We have to show $\text{id} = d^n_j \circ s^{n - 1}_j = d^n_{j + 1} \circ s^{n - 1}_j$, i.e., $$ 1_n = (1_j \star d \star 1_{n - j}) \circ (1_j \star s \star 1_{n - 1 - j}) = (1_{j + 1} \star d \star 1_{n - j - 1}) \circ (1_j \star s \star 1_{n - 1 - j}) $$ This is easily seen to be implied by the first assumption of the lemma. \medskip\noindent We check condition (1)(d). Let $0 < j + 1 < i \leq n$. We have to show $d^n_i \circ s^{n - 1}_j = s^{n - 2}_j \circ d^{n - 1}_{i - 1}$, i.e., $$ (1_i \star d \star 1_{n - i}) \circ (1_j \star s \star 1_{n - 1 - j}) = (1_j \star s \star 1_{n - 2 - j}) \circ (1_{i - 1} \star d \star 1_{n - i}) $$ By the same kind of calculus as in case (1)(a) both sides simplify to $1_j \star s \star 1_{i - j - 2} \star d \star 1_{n - i}$. \medskip\noindent We check condition (1)(e). Let $0 \leq i \leq j \leq n - 1$. We have to show that $s^n_i \circ s^{n - 1}_j = s^n_{j + 1} \circ s^{n - 1}_i$, i.e., $$ (1_i \star s \star 1_{n - i}) \circ (1_j \star s \star 1_{n - 1 - j}) = (1_{j + 1} \star s \star 1_{n - 1 - j}) \circ (1_i \star s \star 1_{n - 1 - i}) $$ By the same kind of calculus as in case (1)(a) this reduces to $$ (s \star 1_{j - i + 1}) \circ (1_{j - i} \star s) = (1_{j - i + 1} \star s) \circ (s \star 1_{j - i}) $$ If $j = i$ this is exactly one of the two assumptions of the lemma. For $j > i$ left and right hand side both reduce to the equality $s \star 1_{j - i - 1} \star s$ by calculations similar to those we did in case (1)(a). \medskip\noindent Finally, in order to show that $d$ defines an augmentation we have to show that $d \circ (1_1 \star d) = d \circ (d \star 1_1)$ which is true because both sides are equal to $d \star d$.