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
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