By Barmak J.A., Minian E.G.

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**Example text**

The inverse image of δ (j) is empty. We see that the map Ψ maps the complement D2 \ ( s Es ) to X (1) , and maps each oval E1 . . , 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.

4. Let XA = α∈A Sα1 . Then π1 (XA ) is a free group with generators ηα , α ∈ A. Proof. Let iα : S 1 −→ XA be an embedding of the corresponding circle. Let ηα ∈ π1 (XA ) be the element given by iα . We prove the following statement. 1. 1 o Any element β ∈ π1 (XA ) may be represented as a finite product of elements ηα , ηα−1 , α ∈ A: β = ηαǫ11 · · · ηαǫss , (14) ǫj = ±1. 2 o The presentation (14) is unique up to cancelation of the elements ηα ηα−1 or ηα−1 ηα . 4. Now we prove 1 o , and we postpone 2 o to the next section.

2 gives a unique lifting of this path to T , such that it starts at f (z). It gives a map Z × I −→ T . This is our homotopy F . 3. Covering spaces and fundamental group. 3. Let p : T −→ X be a covering space, then p∗ : π1 (T, x0 ) −→ π1 (X, x0 ) is a monomorphism (injective). Proof: Let s : I −→ T be a loop, where s(0) = s(1) = x0 . Denote x0 = p(x0 ). Assume that the loop s = p ◦ s : I −→ X is homotopic to zero. Let st : I −→ X be such a homotopy: s0 = s, st (0) = st (1) = x0 , and s1 (I) = x0 .