Get A First Course in Algebraic Topology PDF

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 for you to support self research by way of the novices. New definitions are by way of compatible illustrations and the proofs of the theorems are simply obtainable to the readers. adequate variety of examples were integrated to facilitate transparent knowing of the recommendations. The ebook starts off with the fundamental notions of type, functors and homotopy of constant mappings together with relative homotopy. primary teams of circles and torus were handled besides the basic team of overlaying areas. Simplexes and complexes are awarded intimately and homology theories-simplicial homology and singular homology were thought of in addition to calculations of a few homology teams. The ebook could be best suited to senior graduate and postgraduate scholars of assorted universities and institutes.

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Extra resources for A First Course in Algebraic Topology

Example text

Proof. Since Y is contractible, there exists a continuous mapping F: Y x C -7 Y with F(y, 0) = y and F(y, 1) = y0 for every ye Y, where y0 is a fixed element. Let f: X ~ Y be a continuous mapping. Define';/: Xx C -7 Y by 19 ALGEBRAIC TOPOLOGY 20 ;f(x, t) =F(j(x), t) for x e X. 1, it follows that ;fis continuous. Further =F(f(x), 0) =f(x) ;f (x, 1) = F(j(x), 1) =Yo· ;f (x, 0) and This shows that f is homotopic to g : X ~ Y where g is a constant mapping defined by g(x) =y0 for every x e X. This proves the theorem.

Then X is connected, but not path connected. 6. Let c denote the complex plane and let A= {i}, where i is the imaginary unit B=[O,l]u{(l/n)+iy; neN,O::;y::;l}. and Let X = A u B. We show that X is connected but not path connected. Let for n EN B,1 = [O, 1] u {(l/n) + iy; 0 ::;y::; l}. 3 B,1 is path connected and again by the same theorem, B = neUN Bn is path connected. 4, B is connected. Let U be an open and closed subset of X. We may assume that A c U. If not, we consider the complement of U which is open and closed and contains A.

2. A non-empty open connected subset of R'1 is path connected. 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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