Lemma 14.33.4. Let $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$, $Y$, $d$, $s$, $F$, $G$ be as in Example 14.33.3. Given a transformation of functors $h_0 : G \circ F \to G \circ Y \circ F$ such that

$1_{G \circ F} = (1_ G \star d \star 1_ F) \circ h_0$

Then there is a morphism $h : G \circ F \to G \circ X \circ F$ of simplicial objects such that $\epsilon \circ h = \text{id}$ where $\epsilon : G \circ X \circ F \to G \circ F$ is the augmentation.

Proof. Denote $u_ n : Y = X_0 \to X_ n$ the map of the simplicial object $X$ corresponding to the unique morphism $[n] \to [0]$ in $\Delta$. Set $h_ n : G \circ F \to G \circ X_ n \circ F$ equal to $(1_ G \star u_ n \star 1_ F) \circ h_0$.

For any simplicial object $X = (X_ n)$ in any category $u =(u_ n) : X_0 \to X$ is a morphism from the constant simplicial object on $X_0$ to $X$. Hence $h$ is a morphism of simplicial objects because it is the composition of $1_ G \star u \star 1_ F$ and $h_0$.

Let us check that $\epsilon \circ h = \text{id}$. We compute

$\epsilon _ n \circ (1_ G \star u_ n \star 1_ F) \circ h_0 = \epsilon _0 \circ h_0 = \text{id}$

The first equality because $\epsilon$ is a morphism of simplicial objects and the second equality because $\epsilon _0 = (1_ G \star d \star 1_ F)$ and we can apply the assumption in the statement of the lemma. $\square$

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