By Iain T. Adamson

This paintings goals to provide easy topology in an unconventional method. It offers a evaluation of the elemental definitions including workouts with no suggestions or proofs of the theorems partly 1, after which offers the ideas partly 2, permitting the scholar to check solutions with their very own.

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**Example text**

12. The map deg : π1 (S 1 , 1) → Z is a group isomorphism. All that remains to be checked is that (στ ) (1) = σ (1) + τ (1). Other fundamental groups can now be computed. 13. Let f r{a, b} be the free group on generators a and b, and let Sa1 , Sb2 be two circles. Write Sa1 ∨ Sb1 for the wedge of the two circles joined at x0 . Then there is an isomorphism from f r{a, b} to π1 (Sa1 ∨ Sb1 , x0 ). 14. Recall that there is a double cover p : S n → RP n , and (for n ≥ 2), this cover is universal. The open hemispheres U− = {y ∈ S n | x · y < 0} and U+ = {y ∈ S n | x · y > 0} are even sheets over the open neighbourhood of p(x) given by U = pU− = pU+ .

14. (1) χ(S) = 2; (2) χ(sphere with k cross-caps) = 2 − k; (3) χ(sphere with h handles) = 2 − 2h. So the standard form of any closed surface M is unique, and it is determined by χ(M ) and the orientability of M . 15. (1) abcbca = 2P . (2) abca−1 cb−1 = 3P . (3) abcdef e−1 db−1 af c = 6P . (4) ae−1 a−1 bdb−1 ced−1 c−1 = 2T . 6. Invariance of the characteristic It remains to prove that different triangulations of the same surface cannot give rise to different Euler characteristics. Solution 1: Consider subdivisions of the surface into polygons (not just triangles).

Now each link of v is a simple closed polygon, so there is an unused triangle with v and another vertex (and hence an edge) in common with a used triangle. 4). The resulting figure is a polygonal representation of M . It has n triangles, and the boundary has n + 2 edges in equally labelled pairs. The identifications given by the edge labellings form a symbol: read around the figure, starting anywhere, and use −1 to denote reverse directions. 16, the symbol is −1 −1 e−1 5 e4 e4 e6 e6 e5 . Notice that the triangles in a polygonal representation can always be oriented coherently, but this may not give a coherent orinetation to M : if the symbol contains .