By Czes Kosniowski

ISBN-10: 0521298644

ISBN-13: 9780521298643

This self-contained creation to algebraic topology is appropriate for a couple of topology classes. It contains approximately one region 'general topology' (without its traditional pathologies) and 3 quarters 'algebraic topology' (centred round the primary staff, a quite simply 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 so as to allow the reader to exploit it for self-study in addition to a path booklet. The method is leisurely and a geometrical flavour is clear all through. the various illustrations and over 350 routines will end up important as a instructing relief. This account should be welcomed by means of complex scholars of natural arithmetic at schools and universities.

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Clx(A) the closure of A in X and by Prove that Clx(A). Show that in general Cly(A)*Clx(A). (1) Show that the subset (a,b) of R with the induced topology is homeomorphic to K. ) (g) Let X,Y be topological spaces and let S be a subspace of X. Prove that if f: X V is a continuous map then so is fiS: S f(S). (Ii) Show that the subspaces (1,00), (0,1) of K with the usual topology are homeomorphic. ,0,I) } is homeomorphic to usual topology. ) 26 A first course in algebraic topology - {O} and have the subspace topology of (j) Let with the usual topology.

4 Lemma (I) If S is open in X then the open sets of S in the induced topology are open in X. Induced topology 25 (ii) If S is closed in X then the closed sets of S in the induced topology are closed in X. P-roof Since the proofs of (i) and (ii) are more or less identical we shall only give the proof of(i). Suppose S is open in X and let U be an open subset of S. By definition U = V fl S where V is an open subset of X. But since S is open in X we also have that U = V fl S is open in X. 5 Exercises (a) Show that if Y is a subspace of X, and Z is a subspace of Y, then Z is a subspace of X.

Surjectivity of F is easy to show. To prove that F is continuous we Y consider the natural projections lrX: X X/-x and which are continuous. Clearly Firx = lTy f and since f is continuous we deduce that is continuous and hence F is continuous by the universal mapping pro- Quotient topology (and groups acting on spaces) perty of quotients. The fact that because F' lry iixf' 35 is continuous follows in a similar way R with the equivalence relation Also consider x' if and only if there is an integer n such that x' = x' if and only if there is an integer n R with the equivalence relation x such that x' = n + x.

### A First Course in Algebraic Topology by Czes Kosniowski

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