Lemma 14.33.5. In Situation 14.33.1 the maps

$1_ i \star \epsilon : i \circ X \to i, \quad \text{and}\quad \epsilon \star 1_ F : X \circ F \to F$

are homotopy equivalences.

Proof. Denote $\epsilon _ n : X_ n \to X_{-1}$ the components of the augmentation morphism. We observe that $\epsilon _ n = k^{\star (n + 1)}$, the $(n + 1)$-fold $\star$-composition of $k$. Recall that $t : \text{id}_\mathcal {S} \to i \circ F$ is the adjunction map. We have the morphisms

$t^{\star (n + 1)} \star 1_ i : i \longrightarrow i \circ (F \circ i)^{\circ (n + 1)} = i \circ X_ n$

which are right inverse to $1_ i \star \epsilon _ n$ and the morphisms

$1_ F \star t^{\star (n + 1)} : F \longrightarrow (F \circ i)^{\circ (n + 1)} \circ F = X_ n \circ F$

which are right inverse to $\epsilon _ n \star 1_ F$. These morphisms determine morphisms of simplicial objects $b : i \to i \circ X$ and $c : F \to X \circ F$ (proof omitted). To finish it suffices to construct a homotopy between the morphisms $1, b \circ (1_ i \star \epsilon ) : i \circ X \to i \circ X$ and between the two morphisms $1, c \circ (\epsilon \star 1_ F) : X \circ F \to X \circ F$.

To show the morphisms $b \circ (1_ i \star \epsilon ), 1 : i \circ X \to i \circ X$ are homotopic we have to construct morphisms

$h_{n, j} : i \circ X_ n \to i \circ X_ n$

for $n \geq 0$ and $0 \leq j \leq n + 1$ satisfying the relations described in Lemma 14.26.2. See also Remark 14.26.4. We are forced to set $h_{n, 0} = 1$ and

$h_{n , n + 1} = b_ n \circ (1_ i \star \epsilon _ n) = (t^{\star (n + 1)} \star 1_ i) \circ (1_ i \star k^{\star (n + 1)})$

Thus a logical choice is

$h_{n , j} = (t^{\star (j)} \star 1) \circ (1_ i \star k^{\star (j)} \star 1)$

Here and in the rest of the proof we drop the subscript from $1$ if it is clear by knowing the source and the target of the morphism what this subscript should be. Writing

$i \circ X_ n = i \circ F \circ i \circ \ldots \circ F \circ i$

we can think of the morphism $h_{n, j}$ as collapsing the first $j$ pairs $(F \circ i)$ to $\text{id}_\mathcal {S}$ using $k^{\star (j)}$, then adding a $\text{id}_\mathcal {S}$ in front and expanding this to $j$ pairs $(i \circ F)$ using $t^{\star (j)}$. We have to prove

1. We have $d^ n_ m \circ h_{n, j} = h_{n - 1, j - 1} \circ d^ n_ m$ for $j > m$.

2. We have $d^ n_ m \circ h_{n, j} = h_{n - 1, j} \circ d^ n_ m$ for $j \leq m$.

3. We have $s^ n_ m \circ h_{n, j} = h_{n + 1, j + 1} \circ s^ n_ m$ for $j > m$.

4. We have $s^ n_ m \circ h_{n, j} = h_{n + 1, j} \circ s^ n_ m$ for $j \leq m$.

Recall that $d^ n_ m$ is given by applying $k$ to the $(m + 1)$st pair $(F \circ i)$ in the functor $X_ n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (2) holds (because $k$ does $\star$-commute with $k$, but not with $t$). Similarly, $s^ n_ m$ is given by applying $1_ F \star t \star i_ i$ to the $(m + 1)$st pair $(F \circ i)$ in $X_ n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (4) holds. In the two remaining cases one uses the fact that the compositions in (14.33.2.1) are the identity causes the drop in the index $j$. Some details omitted.

To show the morphisms $1, c \circ (\epsilon \star 1_ F) : X \circ F \to X \circ F$ are homotopic we have to construct morphisms

$h_{n, j} : X_ n \circ F \longrightarrow X_ n \circ F$

for $n \geq 0$ and $0 \leq j \leq n + 1$ satisfying the relations described in Lemma 14.26.2. See also Remark 14.26.4. We are forced to set $h_{n, 0} = 1$ and

$h_{n , n + 1} = c_ n \circ (\epsilon _ n \star 1_ F) = (1_ F \star t^{\star (n + 1)}) \circ (k^{\star (n + 1)} \star 1_ F)$

Thus a logical choice is

$h_{n , j} = (1_ F \star t^{\star (j)} \star 1) \circ (k^{\star (j)} \star 1)$

Here and in the rest of the proof we drop the subscript from $1$ if it is clear by knowing the source and the target of the morphism what this subscript should be. Writing

$X_ n \circ F = F \circ i \circ F \circ \ldots \circ i \circ F$

we can think of the morphism $h_{n, j}$ as collapsing the first $j$ pairs $(F \circ i)$ to $\text{id}_\mathcal {S}$ using $k^{\star (j)}$, then inserting a $\text{id}_\mathcal {S}$ just after the first $F$ and expanding this to $j$ pairs $(i \circ F)$ using $t^{\star (j)}$. We have to prove

1. We have $d^ n_ m \circ h_{n, j} = h_{n - 1, j - 1} \circ d^ n_ m$ for $j > m$.

2. We have $d^ n_ m \circ h_{n, j} = h_{n - 1, j} \circ d^ n_ m$ for $j \leq m$.

3. We have $s^ n_ m \circ h_{n, j} = h_{n + 1, j + 1} \circ s^ n_ m$ for $j > m$.

4. We have $s^ n_ m \circ h_{n, j} = h_{n + 1, j} \circ s^ n_ m$ for $j \leq m$.

Recall that $d^ n_ m$ is given by applying $k$ to the $(m + 1)$st pair $(F \circ i)$ in the functor $X_ n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (2) holds (because $k$ does $\star$-commute with $k$, but not with $t$). Similarly, $s^ n_ m$ is given by applying $1_ F \star t \star i_ i$ to the $(m + 1)$st pair $(F \circ i)$ in $X_ n = (F \circ i)^{\circ (n + 1)}$. Thus it is clear that (4) holds. In the two remaining cases one uses the fact that the compositions in (14.33.2.1) are the identity causes the drop in the index $j$. Some details omitted. $\square$

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