Algebraic Topology Notes(2010 version,complete,175 pages) by Boris Botvinnik

By Boris Botvinnik

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By definition of CW -complex, it is the same as to construct an extension of the map ψ = F (n) ◦ g : (Dn+1 × {0}) ∪ (S n × I) −→ Y to a map of the cylinder ψ ′ : Dn+1 × I −→ Y . Let η : Dn+1 × I −→ (Dn+1 × {0}) ∪ (S n × I) be a projection map of the cylinder Dn+1 × I from a point s which is near and a bit above of the top side Dn+1 × {1} of the cylinder Dn+1 × I , see the Figure below. 0 1 00000 000 00000 000 11111 111 11111 111 0 1 00000 11111 000 111 00000 11111 000 00000 000111 00000 000 11111 111 11111 111 00000 000111 00000 000 11111 111 00000 11111 00011111 111 00000 11111 000 111 00000 11111 000111 111 00000 11111 000 00000 11111 000 111 00000 11111 000 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 11111 000 111 00000 11111 000 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 00000 11111 000111 111 00000 11111 000 111 00000 11111 000 111 00000 11111 000 111 00000 11111 000 111 00000 11111 000 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 00000 000111 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 111 00000 000 00000 000 11111 111 11111 00000 000111 00000 000 11111 111 11111 111 The map η is an identical map on (Dn+1 × {0}) ∪ (S n × I).

Thus (ǫ1 , . . , ǫk ) ∈ E(σ). For each k -frame (v1 , . . , 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.

Dependence of the fundamental group on the base point. 2. Let X be a path-connected space, then π1 (X, x0 ) ∼ = π1 (X, x1 ) for any two points x0 , x1 ∈ X . Proof. Since X is path-connected, there exist a path α : I −→ X , such that α(0) = x0 , α(1) = x1 . We define a homomorphism α# : π1 (X, x0 ) −→ π1 (X, x1 ) as follows. Let [ϕ] ∈ π1 (X, x0 ). We define α# ([ϕ]) = (αϕ)α−1 . 6 It is very easy to check that α# is well−1 defined and is a homomorphism. Moreover, the homomorphism α# : π1 (X, x1 ) −→ π1 (X, x0 ) −1 −1 defined by the formula α# ([ψ]) = [(α ψ)α], gives a homomorphism which is inverse to α# .

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