The Stacks project

Proof. By induction on $n$ we will show that the choice of $N(U_ m)$ in the lemma guarantees that (14.18.1.1) 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 14.3.3.

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 $x'' \in \mathop{\mathrm{Im}}(s_ i)$ 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.

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

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 $x_ i \in \mathop{\mathrm{Ker}}(d_0) \cap \mathop{\mathrm{Ker}}(d_1) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_ i)$ and $z_ i \in \mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{i - 1})$ for $i = n, \ldots , 0$. Here and in the following an empty intersection of kernels indicates the whole space; i.e., the notation $z_0 \in \mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{i - 1})$ when $i = 0$ means $z_0 \in U_ n$ with no restriction.

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

The last equality by the relations of Remark 14.3.3. 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.

The claim implies we can uniquely write

with $x_0 \in N(U_{n + 1})$ and $z_ i \in \mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{i - 1})$. We can reformulate this as saying that we have found a direct sum decomposition

with the property that

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

with $z_{i, 0} \in N(U_ n)$ and $z'_{i, j} \in \mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{j - 1})$. 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

Continuing in this fashion we see that we in the end obtain a decomposition of $x$ as a sum of terms of the form

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

with $0 \leq i_1 \leq i_2 \leq \ldots \leq i_ k \leq n - k + 1$. $\square$