Lemma 14.34.2. In Situation 14.34.1 the system $X = (X_ n, d^ n_ j, s^ n_ j)$ is a simplicial object of $\text{Fun}(\mathcal{A}, \mathcal{A})$ and $\epsilon _0$ defines an augmentation $\epsilon $ from $X$ to the constant simplicial object with value $X_{-1} = \text{id}_\mathcal {A}$.
Proof. Consider $Y = U \circ V : \mathcal{A} \to \mathcal{A}$. We already have the transformation $d : Y = U \circ V \to \text{id}_\mathcal {A}$. Let us denote
\[ s = 1_ U \star \eta \star 1_ V : Y = U \circ \text{id}_\mathcal {S} \circ V \longrightarrow U \circ V \circ U \circ V = Y \circ Y \]
This places us in the situation of Example 14.33.1. It is immediate from the formulas that the $X, d^ n_ i, s^ n_ i$ constructed above and the $X, s^ n_ i, s^ n_ i$ constructed from $Y, d, s$ in Example 14.33.1 agree. Thus, according to Lemma 14.33.2 it suffices to prove that
\[ 1_ Y = (d \star 1_ Y) \circ s = (1_ Y \star d) \circ s \quad \text{and}\quad (s \star 1) \circ s = (1 \star s) \circ s \]
The first equal sign translates into the equality
\[ 1_ U \star 1_ V = (d \star 1_ U \star 1_ V) \circ (1_ U \star \eta \star 1_ V) \]
which holds if we have $1_ U = (d \star 1_ U) \circ (1_ U \star \eta )$ which in turn holds by (14.34.1.1). Similarly for the second equal sign. For the last equation we need to prove
\[ (1_ U \star \eta \star 1_ V \star 1_ U \star 1_ V) \circ (1_ U \star \eta \star 1_ V) = (1_ U \star 1_ V \star 1_ U \star \eta \star 1_ V) \circ (1_ U \star \eta \star 1_ V) \]
For this it suffices to prove $(\eta \star 1_ V \star 1_ U) \circ \eta = (1_ V \star 1_ U \star \eta ) \circ \eta $ which is true because both sides are the same as $\eta \star \eta $. $\square$
Comments (0)
There are also: