By Botvinnik B.

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Ek linearly on one of the cells e2j . We join now a point s0 ∈ S 1 ⊂ D2 with each oval E1 , . . , Ek by paths s1 , . . sk , which do not intersect with each other, see the picture below: Figure 23 We denote by σ1 , . . , σk the loops, going clock-wise around each oval. Then the loop σ going clock-wise along the circle S 1 ⊂ D2 is homotopic in D2 \ t Int(Et ) to the loop: −1 −1 (sk σk s−1 k ) · · · (s2 σ2 s2 )(s1 σ1 s1 ), see Fig. 6. It means that the loop ϕ : S 1 −→ X (2) is homotopic (in X (1) ) to the loop −1 −1 [Ψ ◦ (sk σk s−1 k )] · · · [Ψ ◦ (s2 σ2 s2 )][Ψ ◦ (s1 σ1 s1 )].

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. 2. We define the map f : E(σ1 , . . , σk ) × D −→ E(σ1 , . . , σk , σk+1 ) by the formula f ((v1 , . . , vk ), u) = (v1 , . . , vk , T u) where T is given by (13). We notice that vi , T u = T ǫi , T u = ǫi , u = 0, i = 1, . . , k, and T u, T u = u, u = 1 by definition of T and since T ∈ O(n).

Consider the sphere S 3 ⊂ C2 . Let p be a prime number, and q = 0 mod p.