Download e-book for iPad: A First Course in Algebraic Topology by A. Lahiri

By A. Lahiri

ISBN-10: 1842650033

ISBN-13: 9781842650035

This quantity is an introductory textual content the place the subject material has been provided lucidly to be able to aid self learn by means of the newcomers. New definitions are by means of appropriate illustrations and the proofs of the theorems are simply available to the readers. enough variety of examples were integrated to facilitate transparent figuring out of the ideas. The e-book begins with the fundamental notions of type, functors and homotopy of constant mappings together with relative homotopy. primary teams of circles and torus were taken care of in addition to the elemental staff of protecting areas. Simplexes and complexes are provided intimately and homology theories-simplicial homology and singular homology were thought of in addition to calculations of a few homology teams. The booklet can be best suited to senior graduate and postgraduate scholars of assorted universities and institutes.

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Show that a retract of a Hausdorff space is a closed subset. 7. Show that {O} u { 1} is not a retract of C. 8. Prove that there is a retraction r : B" --+ s•1-I if and only if sn-I is contractible. (Hint : Let F : s11- 1 x c--+ s11- 1 be a homotopy between a constant map and the identity map. Then use the map s•1- 1 x C --+ B" given by (x, t) --+ tx with the fact that F(s11- 1 x {0}) is a single point). 5. Paths _ _ _ _ _ _ __ 1. Definitions Let X be a topological space and C denotes the unit interval [O, 1].

Show that/ ::. g. 2. Let/: S1 --+ X be a continuous map. Show that/is null homotopic if and only if there is a continuous map g: B2 --+ X with/= g I S 1• 3. Let / 0 ::. / 1 (rel A) : X --+ Y and g : Y --+ Z be continuous. Show that g/0 ::. g/1(rel A): X--+ Z. 4. Let/0 ::. / 1 : X--+ Y and g0 ::. g 1 : Y--+ Z. Show that gof0 ::. g 1 / 1 : X--+ Z. 4. Homotopy Type and Retractions _ _ _ _ __ 1. Contractible Spaces If g : X -7 Y is a continuous mapping whose image contains only a single point, then g is a consta_nt mapping.

Proof. Let E be such a set and let p e E. Let F be the subset of E which consists of such points of E each of which can be joined top by a path in E. Considering the null path Ep, we see that Fis non-empty. We show that F is closed and open. Paths 25 To prove that F is open, let q E F c E. Because E is open, there is an E > 0 such that q E D = {x : {x : II q - x II < E } c E. e. any point of D can be joined to q by a path in D. This implies that any point of D can be joined to p by a path in E and so q E D c F.

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A First Course in Algebraic Topology by A. Lahiri

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