Algebraic topology notes by Botvinnik B.

By Botvinnik B.

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Let us choose a cell e0 and for each zero cell e0i choose a path si connecting e0i and e0 (these paths may have nonempty intersections). By Cellular Approximation Theorem we can choose these paths inside 1-skeleton. Now for each path si we glue a 2-disk, identifying a half-circle with si , see the picture: NOTES ON THE COURSE “ALGEBRAIC TOPOLOGY” 41 1 Y e2 111 000 000000 111111 000 111 000000 111111 0000000000000000000000 1111111111111111111111 X 000 111 000000 0000000000000000000000 1111111111111111111111 000111111 111 000000 111111 0000000000000000000 1111111111111111111 0000000000000000000000 1111111111111111111111 1 e1111111111111111111 0000000000000000000 0000000000000000000000 1111111111111111111111 1 0000000000000000000 1111111111111111111 0000000000000000000000 1111111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000000 1111111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000000 0 e1111111111111111111111 0000000000000000000 1111111111111111111 0000000000000000000000 1111111111111111111111 0 0000000000000000000 1111111111111111111 s2 0000000000000000000000 1111111111111111111111 s1 0000000000000000000 1111111111111111111 e02 0000000000000000000 1111111111111111111 0 e 1 X Figure 19 We denote the resulting CW -complex by X .

Proof. For each point x ∈ V¯ there exists a simplex ∆n (x) with a center at x and ∆n (x) ⊂ U . By compactness of V¯ there exist a finite number of simplices ∆n (xi ) covering V¯ . 6 to conclude that a union of finite number of ∆n (xi ) has a finite triangulation. 5. 6. We consider carefully our map ϕ : U −→ D . First we construct the disks d1 , d2 , d3 , d4 inside the disk d with the same center and of radii r/5, 2r/5, 3r/5, 4r/5 respectively, where r is a radius of d. Then we cover V = ϕ−1 (d) by finite number of p-simplexes ∆p (j), such that ∆n (j) ⊂ U .

7 that πn (T 2 ) = 0 for n ≥ 2. 17. Let Kl2 be the Klein bottle. Construct two-folded covering space Kl2 −→ T 2 . Compute πn (Kl2 ) for all n. 11. Let M 2 be a two-dimensional manifold without boundary, M 2 = S 2 , RP2 . Then πn (M 2 ) = 0 for n ≥ 2. 18. 11. Hint: One way is to construct a universal covering space over M 2 ; this universal covering space turns our to be R2 . 8 shows that πn (X) = πn (M 2 ). ). Now it remains to make an argument in a general case. 8. Lens spaces. We conclude with important examples.

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