Definition 20.23.1. Let X be a topological space. Let \mathcal{U} : U = \bigcup _{i \in I} U_ i be an open covering. Let \mathcal{F} be an abelian presheaf on X. The complex \check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F}) is the alternating Čech complex associated to \mathcal{F} and the open covering \mathcal{U}.
20.23 The alternating Čech complex
This section compares the Čech complex with the alternating Čech complex and some related complexes.
Let X be a topological space. Let \mathcal{U} : U = \bigcup _{i \in I} U_ i be an open covering. For p \geq 0 set
We omit the verification that the differential d of Equation (20.9.0.1) maps \check{\mathcal{C}}^ p_{alt}(\mathcal{U}, \mathcal{F}) into \check{\mathcal{C}}^{p + 1}_{alt}(\mathcal{U}, \mathcal{F}).
Hence there is a canonical morphism of complexes
namely the inclusion of the alternating Čech complex into the usual Čech complex.
Suppose our covering \mathcal{U} : U = \bigcup _{i \in I} U_ i comes equipped with a total ordering < on I. In this case, set
This is an abelian group. For s \in \check{\mathcal{C}}_{ord}^ p(\mathcal{U}, \mathcal{F}) we denote s_{i_0\ldots i_ p} its value in \mathcal{F}(U_{i_0\ldots i_ p}). We define
by the formula
for any i_0 < \ldots < i_{p + 1}. Note that this formula is identical to Equation (20.9.0.1). It is straightforward to see that d \circ d = 0. In other words \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F}) is a complex.
Definition 20.23.2. Let X be a topological space. Let \mathcal{U} : U = \bigcup _{i \in I} U_ i be an open covering. Assume given a total ordering on I. Let \mathcal{F} be an abelian presheaf on X. The complex \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F}) is the ordered Čech complex associated to \mathcal{F}, the open covering \mathcal{U} and the given total ordering on I.
This complex is sometimes called the alternating Čech complex. The reason is that there is an obvious comparison map between the ordered Čech complex and the alternating Čech complex. Namely, consider the map
given by the rule
Here \sigma denotes a permutation of \{ 0, \ldots , p\} and \text{sgn}(\sigma ) denotes its sign. The alternating and ordered Čech complexes are often identified in the literature via the map c. Namely we have the following easy lemma.
Lemma 20.23.3. Let X be a topological space. Let \mathcal{U} : U = \bigcup _{i \in I} U_ i be an open covering. Assume I comes equipped with a total ordering. The map c is a morphism of complexes. In fact it induces an isomorphism
of complexes.
Proof. Omitted. \square
There is also a map
which is described by the rule
whenever i_0 < i_1 < \ldots < i_ p.
Lemma 20.23.4. Let X be a topological space. Let \mathcal{U} : U = \bigcup _{i \in I} U_ i be an open covering. Assume I comes equipped with a total ordering. The map \pi : \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) \to \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{F}) is a morphism of complexes. It induces an isomorphism
of complexes which is a left inverse to the morphism c.
Proof. Omitted. \square
Remark 20.23.5. This means that if we have two total orderings <_1 and <_2 on the index set I, then we get an isomorphism of complexes \tau = \pi _2 \circ c_1 : \check{\mathcal{C}}_{ord\text{-}1}(\mathcal{U}, \mathcal{F}) \to \check{\mathcal{C}}_{ord\text{-}2}(\mathcal{U}, \mathcal{F}). It is clear that
where i_0 <_1 i_1 <_1 \ldots <_1 i_ p and i_{\sigma (0)} <_2 i_{\sigma (1)} <_2 \ldots <_2 i_{\sigma (p)}. This is the sense in which the ordered Čech complex is independent of the chosen total ordering.
Lemma 20.23.6. Let X be a topological space. Let \mathcal{U} : U = \bigcup _{i \in I} U_ i be an open covering. Assume I comes equipped with a total ordering. The map c \circ \pi is homotopic to the identity on \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}). In particular the inclusion map \check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F}) \to \check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}) is a homotopy equivalence.
Proof. For any multi-index (i_0, \ldots , i_ p) \in I^{p + 1} there exists a unique permutation \sigma : \{ 0, \ldots , p\} \to \{ 0, \ldots , p\} such that
We denote this permutation \sigma = \sigma ^{i_0 \ldots i_ p}.
For any permutation \sigma : \{ 0, \ldots , p\} \to \{ 0, \ldots , p\} and any a, 0 \leq a \leq p we denote \sigma _ a the unique permutation of \{ 0, \ldots , p\} such that \sigma _ a(j) = \sigma (j) for 0 \leq j < a and such that \sigma _ a(a) < \sigma _ a(a + 1) < \ldots < \sigma _ a(p). So if p = 3 and \sigma , \tau are given by
then we have
It is clear that always \sigma _0 = \text{id} and \sigma _ p = \sigma .
Having introduced this notation we define for s \in \check{\mathcal{C}}^{p + 1}(\mathcal{U}, \mathcal{F}) the element h(s) \in \check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{F}) to be the element with components
where \sigma = \sigma ^{i_0 \ldots i_ p}. The index i_{\sigma (a)} occurs twice in i_{\sigma (0)} \ldots i_{\sigma (a)} i_{\sigma _ a(a)} \ldots i_{\sigma _ a(p)} once in the first group of a + 1 indices and once in the second group of p - a + 1 indices since \sigma _ a(j) = \sigma (a) for some j \geq a by definition of \sigma _ a. Hence the sum makes sense since each of the elements s_{i_{\sigma (0)} \ldots i_{\sigma (a)} i_{\sigma _ a(a)} \ldots i_{\sigma _ a(p)}} is defined over the open U_{i_0 \ldots i_ p}. Note also that for a = 0 we get s_{i_0 \ldots i_ p} and for a = p we get (-1)^ p \text{sign}(\sigma ) s_{i_{\sigma (0)} \ldots i_{\sigma (p)}}.
We claim that
where \sigma = \sigma ^{i_0 \ldots i_ p}. We omit the verification of this claim. (There is a PARI/gp script called first-homotopy.gp in the stacks-project subdirectory scripts which can be used to check finitely many instances of this claim. We wrote this script to make sure the signs are correct.) Write
for the operator given by the rule
The claim above implies that \kappa is a morphism of complexes and that \kappa is homotopic to the identity map of the Čech complex. This does not immediately imply the lemma since the image of the operator \kappa is not the alternating subcomplex. Namely, the image of \kappa is the “semi-alternating” complex \check{\mathcal{C}}_{semi\text{-}alt}^ p(\mathcal{U}, \mathcal{F}) where s is a p-cochain of this complex if and only if
for any (i_0, \ldots , i_ p) \in I^{p + 1} with \sigma = \sigma ^{i_0 \ldots i_ p}. We introduce yet another variant Čech complex, namely the semi-ordered Čech complex defined by
It is easy to see that Equation (20.9.0.1) also defines a differential and hence that we get a complex. It is also clear (analogous to Lemma 20.23.4) that the projection map
is an isomorphism of complexes.
Hence the Lemma follows if we can show that the obvious inclusion map
is a homotopy equivalence. To see this we use the homotopy
We claim that
We omit the verification. (There is a PARI/gp script called second-homotopy.gp in the stacks-project subdirectory scripts which can be used to check finitely many instances of this claim. We wrote this script to make sure the signs are correct.) The claim clearly shows that the composition
of the projection with the natural inclusion is homotopic to the identity map as desired. \square
Lemma 20.23.7. Let X be a topological space. Let \mathcal{F} be an abelian presheaf on X. Let \mathcal{U} : U = \bigcup _{i \in I} U_ i be an open covering. If U_ i = U for some i \in I, then the extended alternating Čech complex
obtained by putting \mathcal{F}(U) in degree -1 with differential given by the canonical map of \mathcal{F}(U) into \check{\mathcal{C}}^0(\mathcal{U}, \mathcal{F}) is homotopy equivalent to 0. Similarly, for any total ordering on I the extended ordered Čech complex
is homotopy equivalent to 0.
Second proof. Since the alternating and ordered Čech complexes are isomorphic it suffices to prove this for the ordered one. We will use standard notation: a cochain s of degree p in the extended ordered Čech complex has the form s = (s_{i_0 \ldots i_ p}) where s_{i_0 \ldots i_ p} is in \mathcal{F}(U_{i_0 \ldots i_ p}) and i_0 < \ldots < i_ p. With this notation we have
Fix an index i \in I with U = U_ i. As homotopy we use the maps
given by the rule
Here j is the unique index such that i_ j < i < i_{j + 1} in the second case; also, since U = U_ i we have the equality
which we can use to make sense of thinking of (-1)^ j s_{i_0 \ldots i_ j i i_{j + 1} \ldots i_ p} as an element of \mathcal{F}(U_{i_0 \ldots i_ p}). We will show by a computation that d h + h d equals the negative of the identity map which finishes the proof. To do this fix s a cochain of degree p and let i_0 < \ldots < i_ p be elements of I.
Case I: i \in \{ i_0, \ldots , i_ p\} . Say i = i_ t. Then we have h(d(s))_{i_0 \ldots i_ p} = 0. On the other hand we have
Thus (dh + hd)(s)_{i_0 \ldots i_ p} = -s_{i_0 \ldots i_ p} as desired.
Case II: i \not\in \{ i_0, \ldots , i_ p\} . Let j be such that i_ j < i < i_{j + 1}. Then we see that
On the other hand we have
Adding these up we obtain (dh + hd)(s)_{i_0 \ldots i_ p} = - s_{i_0 \ldots i_ p} as desired. \square
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