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
with boundary maps d_ n : U_ n \to U_{n - 1} given by the formula
This is a complex because, by the relations listed in Remark 14.3.3, we have
The signs cancel! We denote the associated chain complex s(U). Clearly, the construction is functorial and hence defines a functor
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
and
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
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
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
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
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
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.
Proof. Define D(U)_ n to be the image of
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
This composition is the sum of the maps
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
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
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.
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
is injective. Thus we get a short exact sequence
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
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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