Algebraic topology notes, Edition: draft by Botvinnik B.

By Botvinnik B.

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Consider the following subspace D ⊂ H D= u∈H σk+1 σk+1 : | |u| = 1, ǫj , u = 0, j = 1, . . , k . 12. Prove that D is homeomorphic to the hemisphere of the dimension σk+1 − k − 1. Thus D is a closed cell of dimension σk+1 −k −1. 2. We define the map f : E(σ1 , . . , σk ) × 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).

We notice that it is enough to define a map (n) F1 : X ∪ ((A ∪ X (n) ∪ en+1 ) × I) −→ Y extending F (n) to a single cell en+1 . Let en+1 be a (n + 1)-cell such that en+1 ⊂ X \ A. 34 BORIS BOTVINNIK By induction, the map F (n) is already given on the cylinder (¯ en+1 \en+1 )×I since the boundary n+1 n+1 n+1 (n) n+1 (n+1) ∂e = e¯ \e ⊂ X . Let g : D −→ X be a characteristic map corresponding (n) n+1 to the cell e . We have to define an extension of F1 from the side g(S n ) × I and the bottom base g(Dn+1 ) × {0} to the cylinder g(Dn+1 ) × I .

11. 12. Let Y be n-connected CW -complex, and X be an n-dimensional CW complex. Then the set [X, Y ] consists of a single element. A pair of spaces (X, A) is n-connected if for any k ≤ n and any map of pairs f : (Dk , S k−1 ) −→ (X, A) homotopic to a map g : (Dk , S k−1 ) −→ (X, A) (as a map of pairs) so that g(Dk ) ⊂ A. 11. What does it mean geometrically that a pair (X, A) is 0-connected? 1connected? Give some alternative description. 12. Let (X, A) be an n-connected pair of CW -complexes. Prove that (X, A) is homotopy equivalent to a CW -pair (Y, B) so that B ⊂ Y (n) .

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