Geometry And Topology

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

By Boris Botvinnik

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Extra info for Algebraic Topology Notes(2010 version,complete,175 pages)

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Now let A([γ]) = A([γ ′ ]). Then the loop β = (γ ′ )−1 γ is covered by the loop β = (γ ′ )−1 γ , see Fig. (b). Thus [β] = [(γ ′ )−1 γ] ∈ p∗ (π1 (T, x0 )). γ(1) x0 γ(1) γ x0 γ′ x0 γ′ x0 γ (a) γ γ (b) 54 BORIS BOTVINNIK This proves that A : π1 (X, x0 )/p∗ (π1 (T, x0 )) −→ p−1 (x0 ) is an injection. Clearly A is onto since T is path-connected, and if x ∈ p−1 (x0 ) there exists a path connecting x0 and x which projects to a loop in X . 2. Let p : T −→ X be a covering and x0 , x1 ∈ X . There is one-to-one correspondence p−1 (x0 ) ←→ p−1 (x1 ).

Homotopy groups and covering spaces. 8. Let p : T −→ X be a covering space and n ≥ 2. Then the homomorphism p∗ : πn (T, x0 ) −→ πn (X, x0 ) is an isomorphism. 15. 8. 8 allows us to compute homotopy groups of several important spaces. Actually there are only few spaces where all homotopy groups are known. Believe me or not, here we have at least half of those examples. 9. πn (S 1 ) = Z if n = 1, 0 if n ≥ 2. 8 to the covering space R −−→ S 1 ; of course, one should be able to prove πn (R) = 0 for all n ≥ 0.

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. Let S 1 = {z ∈ C | |z| = 1}. The group S 1 acts freely on the sphere S 2n−1 ⊂ Cn by (z1 , .

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