By Christoph Schweigert
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Additional info for Algebraic Topology [Lecture notes]
Example text
We now consider a relative situation, so let X be a topological space with A, B ⊂ X open in A ∪ B and set U := {A, B}. This is an open covering of A ∪ B ⊂ X. The following diagram 35 of exact sequences combines absolute chains with relative ones: 0 0 G 0 0 Sn (A ∩ B) G Sn (X) G 0 G 0 Δ G Sn (A) ⊕ Sn (B) diff Sn (X) ⊕ Sn (X) G Sn (X, A) ⊕ Sn (X, B) Sn (X, A ∩ B) G Sn hhh ϕ hhhhhh h h h hhh hhhh G S U (A ∪ B) G0 S (X) n h n h hhhh hhhh h h h hh hhhh hhhh G Sn (X) G0 Sn (X, A ∪ B) hhR ψ hhhhhh h h h hhh hhhh G0 (X)/S U (A ∪ B) 0 n 0 S (A ∪ B) R n 0 0 0 Here, ψ is induced by the inclusion ϕ : SnU (A ∪ B) → Sn (A ∪ B), Δ denotes the diagonal map and diff is the difference map.
We inductively construct retractions ρr : X × {0} ∪ (A × I ∪ X r × I) → X × {0} ∪ A × [0, 1] , where ρr+1 extends ρr . Suppose that ρr−1 is given. Then extending to an r-cell of X amounts to extending on Dr × [0, 1] along Dr × {0} ∪ Sr−1 × [0, 1]. As we have seen, this can be done. These maps for all r-cells fit together to a map on the r-skeleton (X × [0, 1])r which, by the weak topology, fit together to a retract X × [0, 1] → X × {0} ∪ A × [0, 1]. 19 52 1. A map p : E → B has the homotopy lifting property, if for any space Y and any homotopy h : Y × [0, 1] → B and any map g : Y → E such that p ◦ g = h0 there exists a map H : Y × I → E with p ◦ H = h such that H(y, 0) = g(y) for all y ∈ Y .
Proof. As products of cells are cells, X ×Y inherits a cell decomposition from its factors. Characteristic maps are products of the characteristic maps for the factors. Closure finitenss follows from σ × τ = σ × τ . We need to ensure that X × Y carries the weak topology. To this end, we need a few auxiliary facts: • For two spaces U, V , let C(U, V ) be the set of all continuous maps from U to V . The topology of C(U, V ) is generated (under finite intersections and arbitrary unions) by the sets V (K, O) := {f ∈ C(U, V )|f (K) ⊂ O} for compact K ⊂ U and open O ⊂ V .