The Stacks project

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

\[ N(U_ m) = \bigcap \nolimits _{i = 0}^{m - 1} \mathop{\mathrm{Ker}}(d^ m_ i). \]

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

\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 $x_ i \in \mathop{\mathrm{Ker}}(d_0) \cap \mathop{\mathrm{Ker}}(d_1) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_ i)$ 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

\[ 0 = d_ j(x_{i - 1}) = d_ j(x_ i) + d_ j(s_ i(z_ i)) = d_ j(x_ i) + s_{i - 1}(d_ j(z_ i)). \]

The last equality by the relations of Remark 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

\[ x = s_0(z_0) + s_1(z_1) + \ldots + s_ n(z_ n) + x_ n \]

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

\[ U_{n + 1} = N(U_{n + 1}) \oplus \bigoplus \nolimits _{i = 0}^{i = n} s_ i\Big(\mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{i - 1})\Big) \]

with the property that

\[ \mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_ j) = N(U_{n + 1}) \oplus \bigoplus \nolimits _{i = j + 1}^{i = n} s_ i\Big(\mathop{\mathrm{Ker}}(d_ n) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{i - 1})\Big) \]

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 $z'_{i, j} \in \mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{j - 1})$. 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

\[ x = x_ n + s_0(z_{0, 0}) + s_1(z_{1, 0}) + \ldots + s_ n(z_{n, 0}) + \sum \nolimits _{0 \leq i \leq j \leq n - 1} s_ i(s_ j(z'_{i, j})). \]

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


Comments (3)

Comment #8822 by Connor Bass on

There's a typo at the end of the sentence which begins as "Here is a procedure for decomposing...". It should be "with ", instead of "".

Comment #8823 by Connor Bass on

Another typo, which appears twice: it should be "" for one decomposition of (which appears towards the middle of the proof) instead of "" and it should be "" instead of "" for another (which appears towards the end of the proof).

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  • 6 comment(s) on Section 14.18: Splitting simplicial objects

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