Lemma 14.33.6. Let $\mathcal{C}$, $Y$, $d$, $s$ be as in Example 14.33.1 satisfying the equations of Lemma 14.33.2. Let $f : \text{id}_\mathcal {C} \to \text{id}_\mathcal {C}$ be an endomorphism of the identity functor. Then $f \star 1_ X, 1_ X \star f : X \to X$ are maps of simplicial objects compatible with $f$ via the augmentation $\epsilon : X \to \text{id}_\mathcal {C}$. Moreover, $f \star 1_ X$ and $1_ X \star f$ are homotopic.

Proof. The map $f \star 1_ X$ is the map with components

$X_ n = \text{id}_\mathcal {C} \circ X_ n \xrightarrow {f \star 1_{X_ n}} \text{id}_\mathcal {C} \circ X_ n = X_ n$

For a transformation $a : F \to G$ of endofunctors of $\mathcal{C}$ we have $a \circ (f \star 1_ F) = f \star a = (f \star 1_ G) \circ a$. Thus $f \star 1_ X$ is indeed a morphism of simplicial objects. Similarly for $1_ X \star f$.

To show the morphisms are homotopic we construct morphisms $h_{n, i} : X_ n \to X_ n$ for $n \geq 0$ and $0 \leq i \leq n + 1$ satisfying the relations described in Lemma 14.26.2. See also Remark 14.26.4. It turns out we can take

$h_{n, i} = 1_ i \star f \star 1_{n + 1 - i}$

where $1_ i$ is the identity transformation on $Y \circ \ldots \circ Y$ as in the proof of Lemma 14.33.2. We have $h_{n, 0} = f \star 1_{X_ n}$ and $h_{n, n + 1} = 1_{X_ n} \star f$ which checks the first condition. In checking the other conditions we use the comments made in the proof of Lemma 14.33.5 about the maps $d^ n_ j$ and $s^ n_ j$.

We check condition (2) of Lemma 14.26.2. Let $i > j$. We have to show

$d^ n_ j \circ (1_ i \star f \star 1_{n + 1 - i}) = (1_{i - 1} \star f \star 1_{n + 1 - i}) \circ d^ n_ j$

Since $i - 1 \geq j$ we can use one of the possible descriptions of $d^ n_ j$ to rewrite the left hand side as

$(d^{i - 1}_ j \star 1_{n - i + 1}) \circ (1_ i \star f \star 1_{n + 1 - i}) = d^{i - 1}_ j \star f \star 1_{n + 1 - i}$

Similarly the right hand side becomes

$(1_{i - 1} \star f \star 1_{n + 1 - i}) \circ (d^{i - 1}_ j \star 1_{n - i + 1}) = d^{i - 1}_ j \star f \star 1_{n + 1 - i}$

Thus we obtain the same result and (2) is checked.

The conditions (3), (4), and (5) of Lemma 14.26.2 are checked in exactly the same manner using the strategy of the proof of Lemma 14.33.5. We omit the details1. $\square$

 When $f$ is invertible it suffices to prove that $(a_ n) = 1_ X$ and $(b_ n) = f^{-1} \star 1_ X \star f$ are homotopic. But this follows from Lemma 14.33.5 because in this case $a = b = 1_{\text{id}_\mathcal {C}}$.

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