Download A First Course in Algebraic Topology by Czes Kosniowski PDF

By Czes Kosniowski

This self-contained creation to algebraic topology is acceptable for a couple of topology classes. It includes approximately one zone 'general topology' (without its ordinary pathologies) and 3 quarters 'algebraic topology' (centred round the basic staff, a with no trouble grasped subject which provides a good suggestion of what algebraic topology is). The ebook has emerged from classes given on the college of Newcastle-upon-Tyne to senior undergraduates and starting postgraduates. it's been written at a degree with a view to permit the reader to exploit it for self-study in addition to a path ebook. The strategy is leisurely and a geometrical flavour is obvious all through. the various illustrations and over 350 routines will end up priceless as a instructing reduction. This account might be welcomed via complicated scholars of natural arithmetic at schools and universities.

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Example text

As a second example first consider the space C { (x,y,z)ER3;x2 +y2 l,IzI< l} with the induced topology. e. M= {{p,-p);pEC}. Since we have a natural surjective map from C to M we can give M the quotient topology; the result is called a Mobius strip or band (sometimes Mobius is spelt Moebius). 28 A first course in algebraic topology Consider the function f: M { R3 given by p,- p } -+ ((x2 - y2) (2+xz), 2xy(2+xz),yz) where p = (x,y,z) E C ç R3. It is not difficult to check that f is injective. 1. 2 below).

B) Prove that the space Y is Hausdorif if and only if the diagonal D = { Y;y1= y2 } in YX Y is a closed subset of YX Y. (c) Let f: X Y be a continuous map. Prove that if Y is Hausdorif then the set { (x1,x2) E X X X; f(x1) = f(x2) } is a closed subset of XXX. (d) Let f: X -+ Y be a map which is continuous, open and onto. Prove that Y is a Hausdorff space if and only if the set (x1,x2) E X X X; f(x1) = f(x2) } is a closed subset of XX X. Let X be a compact Hausdorff space and let Y be a quotient space determined by a map f: X Y.

Proof Suppose that X and Y are compact. Let { W3;j Ci I be an open cover is of the form U (U3 k X VJk) where Uj,k is of X X Y. By definition each kEK X VJ,k;J Ci, kEK} is an open in Xand Vj,k is open in Y. Thus { open cover of X X Y. ,n(x) } x } X Y. ,n(xj) } . is a finite open cover of X X Y. (xj) ç Ui,k X Vj,k c E J ) which covers X X Y. It follows that there is a finite subcover of Conversely if X X Y is compact then X and Y are compact because lrX and iTy are continuous. More generally, of course, if X1 are compact topological spaces is also compact.

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