Lemma 14.33.2. In Example 14.33.1 if

then $X = (X_ n, d^ n_ j, s^ n_ j)$ is a simplicial object in the category of endofunctors of $\mathcal{C}$ and $d : X_0 = Y \to \text{id}_\mathcal {C}$ defines an augmentation.

Lemma 14.33.2. In Example 14.33.1 if

\[ 1_ Y = (d \star 1_ Y) \circ s = (1_ Y \star d) \circ s \quad \text{and}\quad (s \star 1) \circ s = (1 \star s) \circ s \]

then $X = (X_ n, d^ n_ j, s^ n_ j)$ is a simplicial object in the category of endofunctors of $\mathcal{C}$ and $d : X_0 = Y \to \text{id}_\mathcal {C}$ defines an augmentation.

**Proof.**
To see that we obtain a simplicial object we have to check that the relations (1)(a) – (e) of Lemma 14.3.2 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 4.28.2.

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.

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 $1_ i \star d \star 1_{j - i - 1} \star s \star 1_{n - j - 1}$.

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.

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 $1_ j \star s \star 1_{i - j - 2} \star d \star 1_{n - i}$.

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).

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 $d \star d$. $\square$

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