Lemma 14.33.5. Let \mathcal{A}, \mathcal{B}, \mathcal{C}, Y, d, s, F, G be as in Example 14.33.3. Let F' : \mathcal{A} \to \mathcal{C} and G' : \mathcal{C} \to \mathcal{B} be two functors. Let (a_ n) : G \circ X \to G' \circ X be a morphism of simplicial objects compatible via augmentations with a : G \to G'. Let (b_ n) : X \circ F \to X \circ F' be a morphism of simplicial objects compatible via augmentations with b : F \to F'. Then the two maps
a \star (b_ n), (a_ n) \star b : G \circ X \circ F \to G' \circ X \circ F'
are homotopic.
Proof.
To show the morphisms are homotopic we construct morphisms
h_{n, i} : G \circ X_ n \circ F \to G' \circ X_ n \circ F'
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. To satisfy condition (1) of Lemma 14.26.2 we are forced to set h_{n, 0} = a \star b_ n and h_{n , n + 1} = a_ n \star b. Thus a logical choice is
h_{n , i} = a_{i - 1} \star b_{n - i}
for 1 \leq i \leq n. Setting a = a_{-1} and b = b_{-1} we see the displayed formular holds for 0 \leq i \leq n + 1.
Recall that
d^ n_ j = 1_ G \star 1_ j \star d \star 1_{n - j} \star 1_ F
on G \circ X \circ F where we use the notation 1_ a = 1_{Y \circ \ldots \circ Y} introduced in the proof of Lemma 14.33.2. We are going to use below that we can rewrite this as
\begin{align*} d^ n_ j & = d^ j_ j \star 1_{n - j} = d^{j + 1}_ j \star 1_{n - j} = \ldots = d^{n - 1}_ j \star 1_1 \\ & = 1_ j \star d^{n - j}_0 = 1_{j - 1} \star d^{n - j + 1}_1 = \ldots = 1_1 \star d^{n - 1}_{j - 1} \end{align*}
Of course we have the analogous formulae for d^ n_ j on G' \circ X \circ F'.
We check condition (2) of Lemma 14.26.2. Let i > j. We have to show
d^ n_ j \circ (a_{i - 1} \star b_{n - i}) = (a_{i - 2} \star b_{n - 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 (a_{i - 1} \star b_{n - i}) = (d^{i - 1}_ j \circ a_{i - 1}) \star b_{n - i} = (a_{i - 2} \circ d^{i - 1}_ j) \star b_{n - i}
Similarly the right hand side becomes
(a_{i - 2} \star b_{n - i}) \circ (d^{i - 1}_ j \star 1_{n - i + 1}) = (a_{i - 2} \circ d^{i - 1}_ j) \star b_{n - i}
Thus we obtain the same result and (2) is checked.
We check condition (3) of Lemma 14.26.2. Let i \leq j. We have to show
d^ n_ j \circ (a_{i - 1} \star b_{n - i}) = (a_{i - 1} \star b_{n - 1 - i}) \circ d^ n_ j
Since j \geq i we may rewrite the left hand side as
(1_ i \star d^{n - i}_{j - i}) \circ (a_{i - 1} \star b_{n - i}) = a_{i - 1} \star (b_{n - 1 - i} \circ d^{n - i}_{j - i})
A similar manipulation shows this agrees with the right hand side.
Recall that
s^ n_ j = 1_ G \star 1_ j \star s \star 1_{n - j} \star 1_ F
on G \circ X \circ F. We are going to use below that we can rewrite this as
\begin{align*} s^ n_ j & = s^ j_ j \star 1_{n - j} = s^{j + 1}_ j \star 1_{n - j - 1} = \ldots = s^{n - 1}_ j \star 1_1 \\ & = 1_ j \star s^{n - j}_0 = 1_{j - 1} \star s^{n - j + 1}_1 = \ldots = 1_1 \star s^{n - 1}_{j - 1} \end{align*}
Of course we have the analogous formulae for s^ n_ j on G' \circ X \circ F'.
We check condition (4) of Lemma 14.26.2. Let i > j. We have to show
s^ n_ j \circ (a_{i - 1} \star b_{n - i}) = (a_ i \star b_{n - i}) \circ s^ n_ j
Since i - 1 \geq j we can rewrite the left hand side as
(s^{i - 1}_ j \star 1_{n - i + 1}) \circ (a_{i - 1} \star b_{n - i}) = (s^{i - 1}_ j \circ a_{i - 1}) \star b_{n - i} = (a_ i \circ s^{i - 1}_ j) \star b_{n - i}
Similarly the right hand side becomes
(a_ i \star b_{n - i}) \circ (s^{i - 1}_ j \star 1_{n - i + 1}) = (a_ i \circ s^{i - 1}_ j) \star b_{n - i}
as desired.
We check condition (5) of Lemma 14.26.2. Let i \leq j. We have to show
s^ n_ j \circ (a_{i - 1} \star b_{n - i}) = (a_{i - 1} \star b_{n + 1 - i}) \circ s^ n_ j
This equality holds because both sides evaluate to a_{i - 1} \star (s^{n - i}_{j - i} \circ b_{n - i}) = a_{i - 1} \star (b_{n + 1 - i} \circ s^{n - i}_{j - i}) by exactly the same arguments as above.
\square
Comments (0)