The Alexander–Whitney chain map

The formula There is a well-known natural transformation α, which consists of homomorphisms

α(X, Y): C(X × Y) − ; − ; → C(X) ⊗ C(Y)

for all spaces X and Y, where C(X) denotes the singular chain complex of X. In degree n, it is defined on any singular simplex σ: Δ^{n} → X × Y, with coordinates σ_{1}: Δ^{n} → X and σ_{2}: Δ^{n} → Y, by the Alexander–Whitney formula:

α_{n}σ = ∑_{k+l=n} σ_{1} ^{∘} λ_{k}^{n} ⊗ σ_{2} ^{∘} ρ_{l}^{n} in C(X) ⊗ C(Y),  (1)

where λ_{k}^{n}: Δ^{k} ⊂ Δ^{n} and ρ_{l}^{n}: Δ^{l} ⊂ Δ^{n} are the linear maps defined on the vertices by λ_{k}^{n}(e_{i}) = e_{i} (for 0 ≤ i ≤ k) and ρ_{l}^{n}(e_{i}) = e_{n−l+i} (for 0 ≤ i ≤ l). The map λ_{k}^{n} embeds Δ^{k} as the “lowest” k-dimensional face of Δ^{n} and ρ_{l}^{n} embeds Δ^{l} as the “highest” l-dimensional face of Δ^{n}.

Proposition 2 The homomorphisms α_{n} do in fact form an augmented chain map.

Iterated face maps We can easily express the maps λ_{k}^{n} and ρ_{l}^{n} in terms of the standard face inclusions η_{i}: Δ^{n} ⊂ Δ^{n+1} (for 0 ≤ i ≤ n + 1), where η_{i} omits the vertex e_{i} from its image, as follows:

λ_{k}^{n} = η_{n} ^{∘} η_{n−1} ^{∘} … ^{∘} η_{k+2} ^{∘} η_{k+1}: Δ^{k} ⊂ Δ^{n}

and

ρ_{l}^{n} = η_{0} ^{∘} η_{0} ^{∘} … ^{∘} η_{0}: Δ^{l} ⊂ Δ^{n} (with n − l factors).

We need to know how they interact with the maps η_{i}. [The following results may also be proved by induction, using the standard formula

η_{i} ^{∘} η_{j} = η_{j} ^{∘} η_{i−1}: Δ^{n} ⊂ Δ^{n+2} for 0 ≤ j < i ≤ n + 2.  (3)

Both these composites omit e_{j} and e_{i} from their images.]

Lemma 4 We have the following relations:

(a) For maps Δ^{k} → Δ^{n+1} involving λ:

η_{i} ^{∘} λ_{k}^{n} = \{ \begin{matrix}λ_{k+1}^{n+1} ^{∘} η_{i}, & if i ≤ k; \\ λ_{k}^{n+1}, & if i > k. \end{matrix}  (5)

and

λ_{k+1}^{n+1} ^{∘} η_{k+1} = λ_{k}^{n+1}.  (6)

(b) For maps Δ^{l} → Δ^{n+1} involving ρ:

η_{i} ^{∘} ρ_{l}^{n} = \{ \begin{matrix}ρ_{l}^{n+1}, & if i ≤ n − l; \\ ρ_{l+1}^{n+1} ^{∘} η_{i−n+l}, & if i > n − l. \end{matrix}  (7)

and

ρ_{l+1}^{n+1} ^{∘} η_{0} = ρ_{l}^{n+1}.  (8)

Proof In (a), (5) is clear for i > k, as the image of λ_{k}^{n} is spanned by e_{0}, e_{1}, … e_{k}, which does not include e_{i}. If i ≤ k, the image of η_{i} ^{∘} λ_{k}^{n} omits e_{i} and is spanned by e_{0}, e_{1}, …, e_{i−1}, e_{i+1}, … e_{k+1}; we get the same effect by first applying η_{i}: Δ^{k} → Δ^{k+1} and then λ_{k+1}^{n+1}. In (6), η_{k+1} leaves e_{0}, e_{1}, …, e_{k} fixed, and so do λ_{k+1}^{n+1} and λ_{k}^{n+1}.

Part (b) can be proved analogously. Alternatively, it is useful to introduce the inversion linear map u_{n}: Δ^{n} → Δ^{n} (not part of the standard simplicial structure) given by u_{n}(e_{i}) = e_{n−i} for all i. We observe that ρ_{l}^{n} = u_{n}^{∘}λ_{l}^{n}^{∘}u_{l}, u_{n+1}^{∘}η_{i}^{∘}u_{n} = η_{n+1−i}, and u_{n} ^{∘} u_{n} = id; then (b) follows immediately from (a).

Proof of the chain map

Proof of Proposition 2 Given a singular simplex τ: Δ^{n+1} → X ×Y, with coordinates τ_{1}: Δ^{n+1} → X and τ_{2}: Δ^{n+1} → Y, we have to show that ∂α_{n+1}(X, Y) τ = α_{n}(X, Y) ∂τ.

From (1),

∂α_{n+1}τ = ∑_{r+s=n+1} ∂(τ_{1} ^{∘} λ_{r}^{n+1} ⊗ τ_{2} ^{∘} ρ_{s}^{n+1})

= ∑_{r+s=n+1} ∂(τ_{1}^{∘}λ_{r}^{n+1}) ⊗ τ_{2}^{∘}ρ_{s}^{n+1} + ∑_{r+s=n+1}(−1)^{r} τ_{1}^{∘}λ_{r}^{n+1} ⊗ ∂(τ_{2}^{∘}ρ_{s}^{n+1})

= ∑_{r+s=n+1} ∑_{i=0}^{r}(−1)^{i} τ_{1} ^{∘} λ_{r}^{n+1} ^{∘} η_{i} ⊗ τ_{2} ^{∘} ρ_{s}^{n+1}

+ ∑_{r+s=n+1} ∑_{j=0}^{s}(−1)^{r+j} τ_{1} ^{∘} λ_{r}^{n+1} ⊗ τ_{2} ^{∘} ρ_{s}^{n+1} ^{∘} η_{j}

and

α_{n}∂τ = ∑_{m=0}^{n+1}(−1)^{m}α_{n}(τ ^{∘} η_{m}) = ∑_{m=0}^{n+1} ∑_{r+s=n}(−1)^{m} τ_{1} ^{∘} η_{m} ^{∘} λ_{r}^{n} ⊗ τ_{2} ^{∘} η_{m} ^{∘} ρ_{s}^{n}.

We choose k and l such that k + l = n and pick out the terms that lie in C_{k}(X) ⊗ C_{l}(Y). In ∂α_{n+1}τ we find

∑_{i=0}^{k+1}(−1)^{i} U_{i} + ∑_{j=0}^{l+1}(−1)^{k+j} V_{j},

where U_{i} = τ_{1} ^{∘} λ_{k+1}^{n+1} ^{∘} η_{i} ⊗ τ_{2} ^{∘} ρ_{l}^{n+1} and V_{j} = τ_{1} ^{∘} λ_{k}^{n+1} ⊗ τ_{2} ^{∘} ρ_{l+1}^{n+1} ^{∘} η_{j}. In α_{n}∂τ we find ∑_{m=0}^{n+1}(−1)^{m} W_{m}, where W_{m} = τ_{1} ^{∘} η_{m} ^{∘} λ_{k}^{n} ⊗ τ_{2} ^{∘} η_{m} ^{∘} ρ_{l}^{n}.

We note that ∂α_{n+1}τ provides k + l + 4 = n + 4 terms, while α_{n}∂τ provides only n + 2 terms. Equations (5) and (7) show that U_{i} = W_{i} for 0 ≤ i ≤ k, where U_{i} and W_{i} have the same sign. These equations also show that V_{j} = W_{k+j} for 1 ≤ j ≤ l + 1, again with the same sign. The two remaining terms are U_{k+1} and V_{0}; (6) and (8) show that U_{k+1} = V_{0}, and these terms appear with opposite signs and so cancel.

For n = 0, (1) reduces to α_{0}τ = τ_{1} ⊗ τ_{2}, which obviously preserves the augmentation.

Applications We deduce the derivation formula for the cup product,

δ(ϕ ∪ ψ) = (δϕ) ∪ ψ + (−1)^{k}ϕ ∪ (δψ) in C^{k+l+1}(X; R),  (9)

where ϕ ∈ C^{k}(X; R), ψ ∈ C^{l}(X; R) and R is any coefficient ring. Equivalently, we have the derivation formula for the cross product,

δ(ϕ × ψ) = (δϕ) × ψ + (−1)^{k}ϕ × δψ in C^{k+l+1}(X × Y; R),  (10)

where ϕ ∈ C^{k}(X; R) and ψ ∈ C^{l}(Y; R). (In fact, either formula is essentially equivalent to Proposition 2.)

Let us abbreviate (1) to α_{n}σ = ∑_{k} σ′_{k} ⊗ σ′′_{n−k}, where σ′_{k} ∈ C_{k}(X) and σ′′_{n−k} ∈ C_{n−k}(Y). Then given ϕ ∈ C^{k}(X; R) and ψ ∈ C^{l}(Y; R), their cross product ϕ × ψ is defined on a singular (k+l)-simplex σ: Δ^{k+l} → X × Y as

⟨ϕ × ψ, σ⟩ = ⟨ϕ, σ′_{k}⟩⟨ψ, σ′′_{l}⟩.  (11)

Since α is a chain map, we have the commutative diagram

\xymatrix{C_{k+l+1}(X × Y) \ar[d]^{∂} \ar[r]^{α_{k+l+1}} & C(X) ⊗ C(Y) \ar[d]^{∂} \\ C_{k+l}(X × Y) \ar[r]^{α_{k+l}} & C(X) ⊗ C(Y)}

which we evaluate on a singular (k+l+1)-simplex τ: Δ^{k+l+1} → X × Y, and then apply ⟨ϕ, −⟩: C_{k}(X) → R and ⟨ψ, −⟩: C_{l}(Y) → R and multiply. The lower route yields ⟨ϕ×ψ, ∂τ⟩ = ⟨δ(ϕ×ψ), τ⟩. If we take the upper route and recall the boundary ∂ in C(X) ⊗ C(Y), the resulting terms that lie in C_{k}(X) ⊗ C_{l}(Y) are

∂τ′_{k+1} ⊗ τ′′_{l} + (−1)^{k}τ′_{k} ⊗ ∂τ′′_{l+1}.

We apply ϕ and ψ and multiply to obtain

⟨ϕ, ∂τ′_{k+1}⟩⟨ψ, τ′′_{l}⟩ + (−1)^{k}⟨ϕ, τ′_{k}⟩⟨ψ, ∂τ′′_{k+1}⟩
= ⟨δϕ, τ′_{k+1}⟩⟨ψ, τ′′_{l}⟩ + (−1)^{k}⟨ϕ, τ′_{k}⟩⟨δψ, τ′′_{l+1}⟩
= ⟨δϕ × ψ, τ⟩ + (−1)^{k}⟨ϕ × δψ, τ⟩.

Since τ is arbitrary, we have (10).

From this, (9) follows by setting Y = X and writing ϕ ∪ ψ = Δ^{#}(ϕ×ψ), where Δ: X → X × X denotes the diagonal map.