Lemma 14.33.2. In Example 14.33.1 if
then X = (X_ n, d^ n_ j, s^ n_ j) is a simplicial object in the category of endofunctors of \mathcal{C} and d : X_0 = Y \to \text{id}_\mathcal {C} defines an augmentation.
Lemma 14.33.2. In Example 14.33.1 if
then X = (X_ n, d^ n_ j, s^ n_ j) is a simplicial object in the category of endofunctors of \mathcal{C} and d : X_0 = Y \to \text{id}_\mathcal {C} defines an augmentation.
Proof. To see that we obtain a simplicial object we have to check that the relations (1)(a) – (e) of Lemma 14.3.2 are satisfied. We will use the short hand notation
for a \geq 0. With this notation we have
We are repeatedly going to use the rule that for transformations of funtors a, a', b, b' we have (a' \circ a) \star (b' \circ b) = (a' \star b') \circ (a \star b) provided that the \star and \circ compositions in this formula make sense, see Categories, Lemma 4.28.2.
Condition (1)(a) always holds (no conditions needed on d and s). Namely, let 0 \leq i < j \leq n + 1. We have to show that d^ n_ i \circ d^{n + 1}_ j = d^ n_{j - 1} \circ d^{n + 1}_ i, i.e.,
We can rewrite the left hand side as
The second equality is true because d \circ 1_1 = d and 1_{j - i} \circ (1_{j - i - 1} \star d) = 1_{j - i - 1} \star d. A similar computation gives the same result for the right hand side.
We check condition (1)(b). Let 0 \leq i < j \leq n - 1. We have to show that d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_{j - 1} \circ d^{n - 1}_ i, i.e.,
By the same kind of calculus as in case (1)(a) both sides simplify to 1_ i \star d \star 1_{j - i - 1} \star s \star 1_{n - j - 1}.
We check condition (1)(c). Let 0 \leq j \leq n - 1. We have to show \text{id} = d^ n_ j \circ s^{n - 1}_ j = d^ n_{j + 1} \circ s^{n - 1}_ j, i.e.,
This is easily seen to be implied by the first assumption of the lemma.
We check condition (1)(d). Let 0 < j + 1 < i \leq n. We have to show d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_ j \circ d^{n - 1}_{i - 1}, i.e.,
By the same kind of calculus as in case (1)(a) both sides simplify to 1_ j \star s \star 1_{i - j - 2} \star d \star 1_{n - i}.
We check condition (1)(e). Let 0 \leq i \leq j \leq n - 1. We have to show that s^ n_ i \circ s^{n - 1}_ j = s^ n_{j + 1} \circ s^{n - 1}_ i, i.e.,
By the same kind of calculus as in case (1)(a) this reduces to
If j = i this is exactly one of the two assumptions of the lemma. For j > i left and right hand side both reduce to the equality s \star 1_{j - i - 1} \star s by calculations similar to those we did in case (1)(a).
Finally, in order to show that d defines an augmentation we have to show that d \circ (1_1 \star d) = d \circ (d \star 1_1) which is true because both sides are equal to d \star d. \square
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