By induction on $n$ we will show that the choice of $N(U_m)$ in the lemma guarantees that (\ref{equation-splitting}) is an isomorphism for $m \leq n$. This is clear for $n = 0$. In the rest of this proof we are going to drop the superscripts from the maps $d_i$ and $s_i$ in order to improve readability. We will also repeatedly use the relations from Remark \ref{remark-relations}. \medskip\noindent First we make a general remark. For $0 \leq i \leq m$ and $z \in U_m$ we have $d_i(s_i(z)) = z$. Hence we can write any $x \in U_{m + 1}$ uniquely as $x = x' + x''$ with $d_i(x') = 0$ and $x'' \in \Im(s_i)$ by taking $x' = (x - s_i(d_i(x)))$ and $x'' = s_i(d_i(x))$. Moreover, the element $z \in U_m$ such that $x'' = s_i(z)$ is unique because $s_i$ is injective. \medskip\noindent Here is a procedure for decomposing any $x \in U_{n + 1}$. First, write $x = x_0 + s_0(z_0)$ with $d_0(x_0) = 0$. Next, write $x_0 = x_1 + s_1(z_1)$ with $d_n(x_1) = 0$. Continue like this to get \begin{eqnarray*} x & = & x_0 + s_0(z_0), \\ x_0 & = & x_1 + s_1(z_1), \\ x_1 & = & x_2 + s_2(z_2), \\ \ldots & \ldots & \ldots \\ x_{n - 1} & = & x_n + s_n(z_n) \end{eqnarray*} where $d_i(x_i) = 0$ for all $i = n, \ldots, 0$. By our general remark above all of the $x_i$ and $z_i$ are determined uniquely by $x$. We claim that $x_i \in \Ker(d_0) \cap \Ker(d_1) \cap \ldots \cap \Ker(d_i)$ and $z_i \in \Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})$ for $i = n, \ldots, 0$. Here and in the following an empty intersection of kernels indicates the whole space; i.e., the notation $z_0 \in \Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})$ when $i = 0$ means $z_0 \in U_n$ with no restriction. \medskip\noindent We prove this by ascending induction on $i$. It is clear for $i = 0$ by construction of $x_0$ and $z_0$. Let us prove it for $0 < i \leq n$ assuming the result for $i - 1$. First of all we have $d_i(x_i) = 0$ by construction. So pick a $j$ with $0 \leq j < i$. We have $d_j(x_{i - 1}) = 0$ by induction. Hence $$ 0 = d_j(x_{i - 1}) = d_j(x_i) + d_j(s_i(z_i)) = d_j(x_i) + s_{i - 1}(d_j(z_i)). $$ The last equality by the relations of Remark \ref{remark-relations}. These relations also imply that $d_{i - 1}(d_j(x_i)) = d_j(d_i(x_i)) = 0$ because $d_i(x_i)= 0$ by construction. Then the uniqueness in the general remark above shows the equality $0 = x' + x'' = d_j(x_i) + s_{i - 1}(d_j(z_i))$ can only hold if both terms are zero. We conclude that $d_j(x_i) = 0$ and by injectivity of $s_{i - 1}$ we also conclude that $d_j(z_i) = 0$. This proves the claim. \medskip\noindent The claim implies we can uniquely write $$ x = s_0(z_0) + s_1(z_1) + \ldots + s_n(z_n) + x_0 $$ with $x_0 \in N(U_{n + 1})$ and $z_i \in \Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})$. We can reformulate this as saying that we have found a direct sum decomposition $$ U_{n + 1} = N(U_{n + 1}) \oplus \bigoplus\nolimits_{i = 0}^{i = n} s_i\Big(\Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})\Big) $$ with the property that $$ \Ker(d_0) \cap \ldots \cap \Ker(d_j) = N(U_{n + 1}) \oplus \bigoplus\nolimits_{i = j + 1}^{i = n} s_i\Big(\Ker(d_n) \cap \ldots \cap \Ker(d_{i - 1})\Big) $$ for $j = 0, \ldots, n$. The result follows from this statement as follows. Each of the $z_i$ in the expression for $x$ can be written uniquely as $$ z_i = s_i(z'_{i, i}) + \ldots + s_{n - 1}(z'_{i, n - 1}) + z_{i, 0} $$ with $z_{i, 0} \in N(U_n)$ and $z'_{i, j} \in \Ker(d_0) \cap \ldots \cap \Ker(d_{j - 1})$. The first few steps in the decomposition of $z_i$ are zero because $z_i$ already is in the kernel of $d_0, \ldots, d_i$. This in turn uniquely gives $$ x = x_0 + s_0(z_{0, 0}) + s_1(z_{1, 0}) + \ldots + s_n(z_{n, 0}) + \sum\nolimits_{0 \leq i \leq j \leq n - 1} s_i(s_j(z'_{i, j})). $$ Continuing in this fashion we see that we in the end obtain a decomposition of $x$ as a sum of terms of the form $$ s_{i_1} s_{i_2} \ldots s_{i_k} (z) $$ with $0 \leq i_1 \leq i_2 \leq \ldots \leq i_k \leq n - k + 1$ and $z \in N(U_{n + 1 - k})$. This is exactly the required decomposition, because any surjective map $[n + 1] \to [n + 1 - k]$ can be uniquely expressed in the form $$ \sigma^{n - k}_{i_k} \ldots \sigma^{n - 1}_{i_2} \sigma^n_{i_1} $$ with $0 \leq i_1 \leq i_2 \leq \ldots \leq i_k \leq n - k + 1$.