By Collin Donald Roberts

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**Example text**

Let ω ∈ •R (x[1]) be an arbitrary basis element. Write ω = xi1 ∧ · · · ∧ xik for some k. Because the first map in the counterclockwise branch is d = 0, the diagram commutes if and only if the clockwise branch is also the zero map. The output from the clockwise branch is ∂λ (∧k ϕ(ω)) = ∂λ (ϕ(xi1 ) ∧ · · · ∧ ϕ(xik )) k ±ϕ(xi1 ) ∧ · · · ∧ λϕ(xiν ) ∧ · · · ∧ ϕ(xik ) = ν=1 =0 = 0. Since every basis element maps to zero via the clockwise branch, therefore the clockwise branch kills every element. So the diagram commutes as required.

Proof. 3, Corollary to Proposition 2. 3. Let R be a commutative ring and let I ⊂ R be an ideal. Denote RI by R. Let ψ : M → N be an R-module homomorphism, where I annihilates N . Then ψ factors uniquely through R ⊗R M : m ❴ M ψ (1 + I) ⊗ m R ⊗R M 22 6 G ψ N Proof. 3, as follows. There is a well-defined homomorphism of R-modules α : with inverse β : and therefore R ⊗R M ∼ = M R ⊗R M → IM (r + I) ⊗ m → rm + IM M IM R ⊗R M → m + IM → (1 + I) ⊗ m M . IM Now the remark applies to the diagram M ψ M IM 2 G ψ N where for any i ∈ I and m ∈ M , we have ψ(im) = i ψ(m) = 0 ∈N so that IM ⊆ ker ψ.

A diagonal approximation Φ is given by the following diagram, where the maps in higher degrees are determined by the maps in degrees zero, one and two. 53 0y 0y R ⊗y R ˜= ⊗ : x → 1 x →1 G RGev y Ry :x→1 0 Φ0 RGy · 1 x→x x τ →x −1 τ →x −1 (x−1) 1 Φ1 RGy · τ G τ →x τ +τ RGev · (τ σ → N (x )τ σ → N (x )τ τ τ → (x − 1)τ − τ (x − 1) N 2 τ ) y Φ2 RGy · σ G σ→σ +σ +∇N (x ,x )τ τ RGev · (σ ττ y σ ) (x−1) 3 Φ3 RG y · τ σ G RGev · (τ σ τ σ y τσ τ σ ) k+l=i−1 τ τ (σ )(k) (σ )(l) ) N .. y 2i RG ·y σ (i) ..