By Czes Kosniowski

This self-contained creation to algebraic topology is appropriate for a few topology classes. It includes approximately one zone 'general topology' (without its traditional pathologies) and 3 quarters 'algebraic topology' (centred round the primary crew, a with no trouble grasped subject which supplies a good suggestion of what algebraic topology is). The e-book 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 in an effort to allow the reader to exploit it for self-study in addition to a direction ebook. The process is leisurely and a geometrical flavour is clear all through. the various illustrations and over 350 routines will end up precious as a instructing reduction. This account should be welcomed by way of complex scholars of natural arithmetic at schools and universities.

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**Extra resources for A First Course in Algebraic Topology**

**Example text**

2 below). The fact that f_I is continuous is left for the reader to prove; in fact it follows quite easily from results to be proved in Chapter 8. We now state and prove the universal mapping property of quotients. 2 Theorem Let f: X Y be a mapping and suppose that Y has the quotient topology with respect to X. Then a mapping g: Y Z from Yto a topological space Z is continuous if and only if gf is continuous. Proof The function f: X Y is continuous and so if g is continuous then so is the composite gf.

This sometimes gives us a way of thowing that certain spaces are not homeomorphic. 3 (a) (b) (C) (d) Induced topology least intuitively at this stage, the subspaces [0,1] and (0,1) of R are not homeomorphic because if we remove the point 0 from [0,11 then we get (0,1] which (intuitively) is in one piece, whereas if we remove any point from (0,1) then we get (intuitively) two pieces; more precisely it is the dis- joint union of two non-empty open subsets. Now (intuitively) one piece cannot be homeomorphic to two pieces (this would involve cutting, which is not continuous) and so [0,1] cannot be homeomorphic to (0,1).

For x E X and g E G define gx = x(g'1) Show that this defines a (left) action of G on X. Why does the definition gx = xg fail? (b) Let H be a subgroup of a group G. For hE H, g E G define hg to be hg. Show that this defines an action of H on G. (c) Let G be a group and let,Y(G) denote the set of subsets of G. Show that { ghhEU} ,gEG,UE5°(G) defines an action of G on5'°(G). (d) Let G act on X and define the stabilizer of x E X to be the set {gEG;gx=x}. Prove that is a subgroup of G. (e) Let G act on X and define the orbit of x E X to be the subset 1 gx;gEG } of X.