Lemma 14.18.5. Let U be a simplicial abelian group. Then U has a splitting obtained by taking N(U_0) = U_0 and for m \geq 1 taking
Moreover, this splitting is functorial on the category of simplicial abelian groups.
Lemma 14.18.5. Let U be a simplicial abelian group. Then U has a splitting obtained by taking N(U_0) = U_0 and for m \geq 1 taking
Moreover, this splitting is functorial on the category of simplicial abelian groups.
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_1(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_ n \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
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