By L.F. McAuley

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**Sample text**

Obviously Y is contractible. Now note that X/Y ∼ X ∼ X , and the complex X/Y has only one zero cell. Now we use induction. Let us assume that we already have constructed the CW -complex X ′ such that X ′ ∼ X and X ′ has a single zero cell, and it does not have cells of dimensions 1, 2, . . , k − 1, where k ≤ n. Note that a closure of each k -cell of X ′ is a sphere S k by induction. Indeed, an attaching map for every k -cell has to go to X ′(0) . Since X ′ is still k -connected, then the embedding S k −→ X ′ (corresponding to a cell eki ) may be extended to a map Dk+1 −→ X ′ .

Let X be a CW -complex with a single zero-cell e0 = x0 , one-cells e1i , i ∈ I , and twocells e2j , j ∈ J . Then we identify the first skeleton X (1) with i∈I Si1 . The inclusion map Si1 → i∈I Si1 determines an element αi ∈ π1 (X (1) , x0 ). 4 π1 (X (1) , x0 ) is a free group on generators αi , i ∈ I . The characteristic map gj : D2 −→ X of the cell e2j determines attaching map fj : S 1 −→ X (1) which determines an element βj ∈ π1 (X (1) , x0 ) up to conjugation. 7 A linear map I −→ S 1 is given by t → (cos(λt + µ), sin(λt + ν)) for some constants λ, µ, ν .

K ) × D −→ E(σ1 , . . , σk , σk+1 ) by the formula f ((v1 , . . , vk ), u) = (v1 , . . , vk , T u) where T is given by (13). We notice that vi , T u = T ǫi , T u = ǫi , u = 0, i = 1, . . , k, and T u, T u = u, u = 1 by definition of T and since T ∈ O(n). 13. Recall that σk < σk+1 . Prove that T u ∈ H σk+1 if u ∈ D . NOTES ON THE COURSE “ALGEBRAIC TOPOLOGY” 31 The inverse map f −1 : E(σ1 , . . , σk , σk+1 ) −→ E(σ1 , . . , σk ) × D is defined by vj = f −1 vj , j = 1, . . , k, u = f −1 vk+1 = (T −1 vk+1 ) = Tv1 ,ǫ1 ◦ Tv2 ,ǫ2 ◦ · · · · · · ◦ Tvk ,ǫk (vk+1 ) ∈ D.