Lemma 14.23.1. The functor $s$ is exact.
14.23 Simplicial objects and chain complexes
Let $\mathcal{A}$ be an abelian category. See Homology, Section 12.13 for conventions and notation regarding chain complexes. Let $U$ be a simplicial object of $\mathcal{A}$. The associated chain complex $s(U)$ of $U$, sometimes called the Moore complex, is the chain complex
\[ \ldots \to U_2 \to U_1 \to U_0 \to 0 \to 0 \to \ldots \]
with boundary maps $d_ n : U_ n \to U_{n - 1}$ given by the formula
\[ d_ n = \sum \nolimits _{i = 0}^ n (-1)^ i d^ n_ i. \]
This is a complex because, by the relations listed in Remark 14.3.3, we have
\begin{eqnarray*} d_ n \circ d_{n + 1} & = & (\sum \nolimits _{i = 0}^ n (-1)^ i d^ n_ i) \circ (\sum \nolimits _{j = 0}^{n + 1} (-1)^ j d^{n + 1}_ j) \\ & = & \sum \nolimits _{0 \leq i < j \leq n + 1} (-1)^{i + j} d^ n_{j - 1} \circ d^{n + 1}_ i + \sum \nolimits _{n \geq i \geq j \geq 0} (-1)^{i + j} d^ n_ i \circ d^{n + 1}_ j \\ & = & 0. \end{eqnarray*}
The signs cancel! We denote the associated chain complex $s(U)$. Clearly, the construction is functorial and hence defines a functor
\[ s : \text{Simp}(\mathcal{A}) \longrightarrow \text{Ch}_{\geq 0}(\mathcal{A}). \]
Thus we have the confusing but correct formula $s(U)_ n = U_ n$.
Proof. Clear from Lemma 14.22.1. $\square$
Lemma 14.23.2. Let $\mathcal{A}$ be an abelian category. Let $A$ be an object of $\mathcal{A}$ and let $k$ be an integer. Let $E$ be the object described in Lemma 14.22.4. Then the complex $s(E)$ is acyclic.
Proof. For a morphism $\alpha : [n] \to [k + 1]$ we define $\alpha ' : [n + 1] \to [k + 1]$ to be the map such that $\alpha '|_{[n]} = \alpha $ and $\alpha '(n + 1) = k + 1$. Note that if the image of $\alpha $ is $[k]$ or $[k + 1]$, then the image of $\alpha '$ is $[k + 1]$. Consider the family of maps $h_ n : E_ n \to E_{n + 1}$ which maps the summand corresponding to $\alpha $ to the summand corresponding to $\alpha '$ via the identity on $A$. Let us compute $d_{n + 1} \circ h_ n - h_{n - 1} \circ d_ n$. We will first do this in case the category $\mathcal{A}$ is the category of abelian groups. Let us use the notation $x_\alpha $ to indicate the element $x \in A$ in the summand of $E_ n$ corresponding to the map $\alpha $ occurring in the index set. Let us also adopt the convention that $x_\alpha $ designates the zero element of $E_ n$ whenever $\mathop{\mathrm{Im}}(\alpha )$ is not $[k]$ or $[k + 1]$. With these conventions we see that
\[ d_{n + 1}(h_ n(x_\alpha )) = \sum \nolimits _{i = 0}^{n + 1} (-1)^ i x_{\alpha ' \circ \delta ^{n + 1}_ i} \]
and
\[ h_{n - 1}(d_ n(x_\alpha )) = \sum \nolimits _{i = 0}^ n (-1)^ i x_{(\alpha \circ \delta _ i^ n)'} \]
It is easy to see that $\alpha ' \circ \delta ^{n + 1}_ i = (\alpha \circ \delta _ i^ n)'$ for $i = 0, \ldots , n$. It is also easy to see that $\alpha ' \circ \delta ^{n + 1}_{n + 1} = \alpha $. Thus we see that
\[ (d_{n + 1} \circ h_ n - h_{n - 1} \circ d_ n)(x_\alpha ) = (-1)^{n + 1} x_\alpha \]
These identities continue to hold if $\mathcal{A}$ is any abelian category because they hold in the simplicial abelian group $[n] \mapsto \mathop{\mathrm{Hom}}\nolimits (A, E_ n)$; details left to the reader. We conclude that the identity map on $E$ is homotopic to zero, with homotopy given by the system of maps $h'_ n = (-1)^{n + 1}h_ n : E_ n \to E_{n + 1}$. Hence we see that $E$ is acyclic, for example by Homology, Lemma 12.13.5. $\square$
Lemma 14.23.3. Let $\mathcal{A}$ be an abelian category. Let $A$ be an object of $\mathcal{A}$ and let $k$ be an integer. We have $H_ i(s(K(A, k))) = A$ if $i = k$ and $0$ else.
Proof. First, let us prove this if $k = 0$. In this case we have $K(A, 0)_ n = A$ for all $n$. Furthermore, all the maps in this simplicial abelian group are $\text{id}_ A$, in other words $K(A, 0)$ is the constant simplicial object with value $A$. The boundary maps $d_ n = \sum _{i = 0}^ n (-1)^ i \text{id}_ A = 0$ if $n$ odd and $ = \text{id}_ A$ if $n$ is even. Thus $s(K(A, 0))$ looks like this
\[ \ldots \to A \xrightarrow {0} A \xrightarrow {1} A \xrightarrow {0} A \to 0 \]
and the result is clear.
Next, we prove the result for all $k$ by induction. Given the result for $k$ consider the short exact sequence
\[ 0 \to K(A, k) \to E \to K(A, k + 1) \to 0 \]
from Lemma 14.22.4. By Lemma 14.22.1 the associated sequence of chain complexes is exact. By Lemma 14.23.2 we see that $s(E)$ is acyclic. Hence the result for $k + 1$ follows from the long exact sequence of homology, see Homology, Lemma 12.13.6. $\square$
There is a second chain complex we can associate to a simplicial object of $\mathcal{A}$. Recall that by Lemma 14.18.6 any simplicial object $U$ of $\mathcal{A}$ is canonically split with $N(U_ m) = \bigcap _{i = 0}^{m - 1} \mathop{\mathrm{Ker}}(d^ m_ i)$. We define the normalized chain complex $N(U)$ to be the chain complex
\[ \ldots \to N(U_2) \to N(U_1) \to N(U_0) \to 0 \to 0 \to \ldots \]
with boundary map $d_ n : N(U_ n) \to N(U_{n - 1})$ given by the restriction of $(-1)^ nd^ n_ n$ to the direct summand $N(U_ n)$ of $U_ n$. Note that Lemma 14.18.8 implies that $d^ n_ n(N(U_ n)) \subset N(U_{n - 1})$. It is a complex because $d^ n_ n \circ d^{n + 1}_{n + 1} = d^ n_ n \circ d^{n + 1}_ n$ and $d^{n + 1}_ n$ is zero on $N(U_{n + 1})$ by definition. Thus we obtain a second functor
\[ N : \text{Simp}(\mathcal{A}) \longrightarrow \text{Ch}_{\geq 0}(\mathcal{A}). \]
Here is the reason for the sign in the differential.
Lemma 14.23.4. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object of $\mathcal{A}$. The canonical map $N(U_ n) \to U_ n$ gives rise to a morphism of complexes $N(U) \to s(U)$.
Proof. This is clear because the differential on $s(U)_ n = U_ n$ is $\sum (-1)^ i d^ n_ i$ and the maps $d^ n_ i$, $i < n$ are zero on $N(U_ n)$, whereas the restriction of $(-1)^ nd^ n_ n$ is the boundary map of $N(U)$ by definition. $\square$
Lemma 14.23.5. Let $\mathcal{A}$ be an abelian category. Let $A$ be an object of $\mathcal{A}$ and let $k$ be an integer. We have $N(K(A, k))_ i = A$ if $i = k$ and $0$ else.
Proof. It is clear that $N(K(A, k))_ i = 0$ when $i < k$ because $K(A, k)_ i = 0$ in that case. It is clear that $N(K(A, k))_ k = A$ since $K(A, k)_{k - 1} = 0$ and $K(A, k)_ k = A$. For $i > k$ we have $N(K(A, k))_ i = 0$ by Lemma 14.21.9 and the definition of $K(A, k)$, see Definition 14.22.3. $\square$
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, for some complex $D(U)$. The construction $U \mapsto D(U)$ is functorial.
\[ s(U) = N(U) \oplus D(U) \]
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 $N(U_{n - 1})$. $\square$
Remark 14.23.7. In the situation of Lemma 14.23.6 the subcomplex $D(U) \subset s(U)$ can also be defined as the subcomplex with terms Namely, since $U_ m$ is the direct sum of the subobject $N(U_ m)$ and the images of $N(U_ k)$ for surjections $[m] \to [k]$ with $k < m$ this is clearly the same as the definition of $D(U)_ n$ given in the proof of Lemma 14.23.6. Thus we see that if $U$ is a simplicial abelian group, then elements of $D(U)_ n$ are exactly the sums of degenerate $n$-simplices.
\[ D(U)_ n = \mathop{\mathrm{Im}}\left( \bigoplus \nolimits _{\varphi : [n] \to [m]\text{ surjective}, \ m < n} U_ m \xrightarrow {\bigoplus U(\varphi )} U_ n\right) \]
Lemma 14.23.8. The functor $N$ is exact.
Proof. By Lemma 14.23.1 and the functorial decomposition of Lemma 14.23.6. $\square$
Lemma 14.23.9. Let $\mathcal{A}$ be an abelian category. Let $V$ be a simplicial object of $\mathcal{A}$. The canonical morphism of chain complexes $N(V) \to s(V)$ is a quasi-isomorphism. In other words, the complex $D(V)$ of Lemma 14.23.6 is acyclic.
Proof. Note that the result holds for $K(A, k)$ for any object $A$ and any $k \geq 0$, by Lemmas 14.23.3 and 14.23.5. Consider the hypothesis $IH_{n, m}$: for all $V$ such that $V_ j = 0$ for $j \leq m$ and all $i \leq n$ the map $N(V) \to s(V)$ induces an isomorphism $H_ i(N(V)) \to H_ i(s(V))$.
To start of the induction, note that $IH_{n, n}$ is trivially true, because in that case $N(V)_ n = 0$ and $s(V)_ n = 0$.
Assume $IH_{n, m}$, with $m \leq n$. Pick a simplicial object $V$ such that $V_ j = 0$ for $j < m$. By Lemma 14.22.2 and Definition 14.22.3 we have $K(V_ m, m) = i_{m!} \text{sk}_ mV$. By Lemma 14.21.10 the natural morphism
\[ K(V_ m, m) = i_{m!} \text{sk}_ mV \to V \]
is injective. Thus we get a short exact sequence
\[ 0 \to K(V_ m, m) \to V \to W \to 0 \]
for some $W$ with $W_ i = 0$ for $i = 0, \ldots , m$. This short exact sequence induces a morphism of short exact sequence of associated complexes
\[ \xymatrix{ 0 \ar[r] & N(K(V_ m, m)) \ar[r] \ar[d] & N(V) \ar[r] \ar[d] & N(W) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(K(V_ m, m)) \ar[r] & s(V) \ar[r] & s(W) \ar[r] & 0 } \]
see Lemmas 14.23.1 and 14.23.8. Hence we deduce the result for $V$ from the result on the ends. $\square$
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