By Czes Kosniowski

This self-contained creation to algebraic topology is acceptable for a few topology classes. It includes approximately one area 'general topology' (without its ordinary pathologies) and 3 quarters 'algebraic topology' (centred round the basic workforce, a comfortably grasped subject which supplies a good suggestion of what algebraic topology is). The publication has emerged from classes given on the collage of Newcastle-upon-Tyne to senior undergraduates and starting postgraduates. it's been written at a degree that allows you to let the reader to exploit it for self-study in addition to a direction publication. The procedure is leisurely and a geometrical flavour is obvious all through. the numerous illustrations and over 350 workouts will turn out helpful as a educating relief. This account should be welcomed through complicated scholars of natural arithmetic at faculties and universities.

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Prove that g is an open mapping if and only if gir is an open mapping. (b) Let X be a G-space with G fInite. Prove that the natural projection ir: X X/G is a closed mapping. (c) Suppose X is a G-space and H is a normal subgroup of G. Show that X/H is a (G/H)-space and that (X/H)/(G/H) X/G. 6 Product spaces Our final general method of constructing new topological spaces from old ones is through the direct product. Recall that the direct product X X Y of two sets X,Y is the set of ordered pairs (x,y) with x E X and y E Y.

For each a E A there is a pair of disjoint open sets Ua, Va with x in Ua and a in Va. ,Va(n) } which covers A. fl Ua(n) is an open set containing x which is disjoint from each of the Va(j) and hence U c X - A. Thus each point x E X - A has an open set containing it which is contained in X- A,whichmeansthatX- AisopenandAisclosed. 7 leads to an important result. 8 Theorem Suppose that f: X -+ Y is a continuous map from a compact space X Hausdorff spaces to a Hausdorff space Y. Then f is a homeomorphism if and only 1ff is bijective.

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. In particular the unit then the product X1 X X2 X ... X is said to be bounded if there is a is compact.