Introduction to Topology (2nd Edition) (Dover Books on by Theodore W. Gamelin, Robert Everist Greene

By Theodore W. Gamelin, Robert Everist Greene

This quantity explains nontrivial purposes of metric area topology to research, essentially constructing their courting. additionally, subject matters from basic algebraic topology specialise in concrete effects with minimum algebraic formalism. chapters contemplate metric house and point-set topology; the different 2 chapters discuss algebraic topological material. Includes workouts, chosen solutions, and fifty one illustrations. 1983 variation.

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Additional info for Introduction to Topology (2nd Edition) (Dover Books on Mathematics)

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Notice that in this case the covering space has trivial fundamental group: if this is the case, the covering is called universal. (c) A triple covering. The following diagram describes a triple covering of a wedge of two circles. Study the picture carefully – the arrows ensure that pre–images of open sets are open sets. (d) A countable universal covering of the wedge of two circles. Using the same notational conventions, the following diagram gives a cover of the wedge of two circles. Notice that each point now has countably many pre-images.

13) Let D denote the polygonal region representing M but without any edges identified. By a simple induction, we have that χ(D) = 1. 6. INVARIANCE OF THE CHARACTERISTIC 52 On the other hand, each side of M that does not use a new letter appears twice and generates one new vertex, so α0 (M ) − α0 (D) = m + 12 r − (n + r) = m − n − 12 r. The edges of M are glued in pairs, so that α1 (M ) − α1 (D) = 12 (−n − r). Finally there are equal numbers of triangles, so α2 (M ) = α2 (D). Adding up we get χ(M ) − χ(D) = m − 12 n, so χ(M ) = m − 12 n + 1.

This is proved by a simple counting argument. 8 are all distinct. Proof. 13) Let D denote the polygonal region representing M but without any edges identified. By a simple induction, we have that χ(D) = 1. 6. INVARIANCE OF THE CHARACTERISTIC 52 On the other hand, each side of M that does not use a new letter appears twice and generates one new vertex, so α0 (M ) − α0 (D) = m + 12 r − (n + r) = m − n − 12 r. The edges of M are glued in pairs, so that α1 (M ) − α1 (D) = 12 (−n − r). Finally there are equal numbers of triangles, so α2 (M ) = α2 (D).

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