Download PDF by Botvinnik B.: Algebraic topology notes

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By Botvinnik B.

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The CW -complex X has the same cells as X and new cells e1i , e2i (the top half-circles and interior of 2-disks). A boundary of each cell e2i belongs to the first skeleton since the paths si are in the first skeleton. Clearly the complex X is a deformational retract of X (one can deform each cell e2i to the path si ). Let Y be a closure of the union i e1i . 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, .

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) . NOTES ON THE COURSE “ALGEBRAIC TOPOLOGY” 43 6. 1. General definitions. Here we define the homotopy groups πn (X) for all n ≥ 1 and examine their basic properties. Let (X, x0 ) be a pointed space, and (S n , s0 ) be a pointed sphere. We have defined the set [S n , X] as a set of homotopy classes of maps f : S n −→ X , such that f (s0 ) = x0 , and homotopy between maps should preserve this property.

Vk ) ∈ E(σ) consider the transformation: (13) T = Tǫk ,vk ◦ Tǫk−1 ,vk−1 ◦ · · · · · · ◦ Tǫ1 ,v1 : Rn −→ Rn σi First we notice that vi = −ǫi since vi ∈ H . Thus the transformations Tǫi ,vi are well-defined. 11. Prove that the transformation T takes the k -frame (ǫ1 , . . , ǫk ) to the frame (v1 , . . , vk ). 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.

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Algebraic topology notes by Botvinnik B.

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