**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.
$\square$

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