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 \ref{lemma-relations-homotopy}. See also Remark \ref{remark-homotopy-better}. To satisfy condition (1) of Lemma \ref{lemma-relations-homotopy} 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 $0 \leq i \leq n + 1$. \medskip\noindent 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 \ref{lemma-godement}. 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 $G' \circ X \circ F'$. \medskip\noindent We check condition (2) of Lemma \ref{lemma-relations-homotopy}. 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. \medskip\noindent We check condition (3) of Lemma \ref{lemma-relations-homotopy}. 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. \medskip\noindent 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 $G' \circ X \circ F'$. \medskip\noindent We check condition (4) of Lemma \ref{lemma-relations-homotopy}. 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. \medskip\noindent We check condition (5) of Lemma \ref{lemma-relations-homotopy}. 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.