By Ballico E.

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**Sample text**

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, .

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# . The rest of the proof is left to you. Perhaps the isomorphism α# depends on α. Let β be the other path, β(0) = x0 , β(1) = x1 . Let γ = βα−1 which defines an element [γ] ∈ π1 (X, x1 ). 5. Prove that β# = [γ]α# [γ]−1 . 6. Let f : X −→ Y be a homotopy equivalence, and x0 ∈ X . Prove that f∗ : π1 (X) −→ π1 (Y, f (x0 )) is an isomorphism. 4. Fundamental group of circle.

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# . The rest of the proof is left to you. Perhaps the isomorphism α# depends on α. Let β be the other path, β(0) = x0 , β(1) = x1 . Let γ = βα−1 which defines an element [γ] ∈ π1 (X, x1 ). 5. Prove that β# = [γ]α# [γ]−1 .