Lemma 14.23.6. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object of $\mathcal{A}$. The canonical morphism of chain complexes $N(U) \to s(U)$ is split. In fact,

\[ s(U) = N(U) \oplus D(U) \]

for some complex $D(U)$. The construction $U \mapsto D(U)$ is functorial.

**Proof.**
Define $D(U)_ n$ to be the image of

\[ \bigoplus \nolimits _{\varphi : [n] \to [m]\text{ surjective}, \ m < n} N(U_ m) \xrightarrow {\bigoplus U(\varphi )} U_ n \]

which is a subobject of $U_ n$ complementary to $N(U_ n)$ according to Lemma 14.18.6 and Definition 14.18.1. We show that $D(U)$ is a subcomplex. Pick a surjective map $\varphi : [n] \to [m]$ with $m < n$ and consider the composition

\[ N(U_ m) \xrightarrow {U(\varphi )} U_ n \xrightarrow {d_ n} U_{n - 1}. \]

This composition is the sum of the maps

\[ N(U_ m) \xrightarrow {U(\varphi \circ \delta ^ n_ i)} U_{n - 1} \]

with sign $(-1)^ i$, $i = 0, \ldots , n$.

First we will prove by ascending induction on $m$, $0 \leq m < n - 1$ that all the maps $U(\varphi \circ \delta ^ n_ i)$ map $N(U_ m)$ into $D(U)_{n - 1}$. (The case $m = n - 1$ is treated below.) Whenever the map $\varphi \circ \delta ^ n_ i : [n - 1] \to [m]$ is surjective then the image of $N(U_ m)$ under $U(\varphi \circ \delta ^ n_ i)$ is contained in $D(U)_{n - 1}$ by definition. If $\varphi \circ \delta ^ n_ i : [n - 1] \to [m]$ is not surjective, set $j = \varphi (i)$ and observe that $i$ is the unique index whose image under $\varphi $ is $j$. We may write $\varphi \circ \delta ^ n_ i = \delta ^ m_ j \circ \psi \circ \delta ^ n_ i$ for some $\psi : [n - 1] \to [m - 1]$. Hence $U(\varphi \circ \delta ^ n_ i) = U(\psi \circ \delta ^ n_ i) \circ d^ m_ j$ which is zero on $N(U_ m)$ unless $j = m$. If $j = m$, then $d^ m_ m(N(U_ m)) \subset N(U_{m - 1})$ and hence $U(\varphi \circ \delta ^ n_ i)(N(U_ m)) \subset U(\psi \circ \delta ^ n_ i)(N(U_{m - 1}))$ and we win by induction hypothesis.

To finish proving that $D(U)$ is a subcomplex we still have to deal with the composition

\[ N(U_ m) \xrightarrow {U(\varphi )} U_ n \xrightarrow {d_ n} U_{n - 1}. \]

in case $m = n - 1$. In this case $\varphi = \sigma ^{n - 1}_ j$ for some $0 \leq j \leq n - 1$ and $U(\varphi ) = s^{n - 1}_ j$. Thus the composition is given by the sum

\[ \sum (-1)^ i d^ n_ i \circ s^{n - 1}_ j \]

Recall from Remark 14.3.3 that $d^ n_ j \circ s^{n - 1}_ j = d^ n_{j + 1} \circ s^{n - 1}_ j = \text{id}$ and these drop out because the corresponding terms have opposite signs. The map $d^ n_ n \circ s^{n - 1}_ j$, if $j < n - 1$, is equal to $s^{n - 2}_ j \circ d^{n - 1}_{n - 1}$. Since $d^{n - 1}_{n - 1}$ maps $N(U_{n - 1})$ into $N(U_{n - 2})$, we see that the image $d^ n_ n ( s^{n - 1}_ j (N(U_{n - 1}))$ is contained in $s^{n - 2}_ j(N(U_{n - 2}))$ which is contained in $D(U_{n - 1})$ by definition. For all other combinations of $(i, j)$ we have either $d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_{j - 1} \circ d^{n - 1}_ i$ (if $i < j$), or $d^ n_ i \circ s^{n - 1}_ j = s^{n - 2}_ j \circ d^{n - 1}_{i - 1}$ (if $n > i > j + 1$) and in these cases the map is zero because of the definition of $N(U_{n - 1})$.
$\square$

## Comments (1)

Comment #2358 by Simon Pepin Lehalleur on