By Morgan J.W., Lamberson P.J.

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You can easily check that this relation makes O a poset. To see that it is directed, given two open covers {Uα }α∈A and {Vβ }β∈B , the open cover {Uα ∩ Vβ }(α,β)∈A×B refines both of them. ˇ ∗ (X; o), and if o ≤ o′ we have a map r ∗ ′ : For any open cover o ∈ O we have H o,o ˇ ∗ (X; o) → H ˇ ∗ (X; o′ ). Furthermore, if o ≤ o′ ≤ o′′ are open covers of X, then r ∗′ ′′ ◦r ∗ ′ = H o,o o ,o ∗ . Thus, {H ˇ ∗ (X, o), r ∗ ′ }o∈O is a directed system of graded abelian groups and graded ro,o ′′ o,o group homomorphisms.

Then we define another chain complex, the mapping cylindar, (M φ)∗ by (M φ)∗ = C∗ [1] ⊕ D∗ with boundary map ∂ : (M φ)n = Cn−1 ⊕ Dn → (M φ)n−1 = Cn−2 ⊕ Dn−1 46 given by ∂(cn−1 , dn ) = (∂ C cn−1 , ∂ D dn + (−1)n φn−1 (cn−1 )). We can express this as a matrix: ∂= ∂nC 0 (−1)n φn−1 ∂nD Then to check that ∂ 2 = 0, ∂2 = ∂C 0 n (−1) φ ∂ D ∂C 0 n−1 (−1) φ ∂D = (∂ C )2 0 D ⋆ (∂ )2 where, ⋆ = (−1)n ∂ D φ + (−1)n−1 ∂ C = (−1)n [φ∂ C − ∂ D φ] = 0 since φ is a chain map. 2, that this map is in fact (−1)n φ∗ : Hn−1 (C∗ ) → Hn−1 (D∗ ), and is thus an isomorphism.

Proof. Let φ ∈ Cˇ k (X; {Uα }). Then we define r ∗ φ by, r ∗ φ(Vβ(0) , . . , Vβ(k) ) = φ(Urβ(0) , . . , Urβ(k) ) assuming that Vβ(0) ∩ . . ∩ Vβ(k) = ∅, in which case of course r ∗ φ(Vβ(0) , . . , Vβ(k) ) = 0. Then r ∗ φ ∈ Cˇ ∗ (X; {Vβ }). We claim that this defines a cochain map. We have, k+1 (−1)i r ∗ φ(Vβ(0) , . . , Vβ(i) , . . , Vβ(k+1) ) δ(δ(r ∗ φ))((Vβ(0) , . . , Vβ(k) ) = i=0 k+1 (−1)i φ(Urβ(0) , . . , Urβ(i) , . . , Urβ(k+1) ) = i=0 = δφ(Urβ(0) , . . , Urβ(k) ) = r ∗ δφ((Vβ(0) , .

### Algebraic topology by Morgan J.W., Lamberson P.J.

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