A First Course in Algebraic Topology by A. Lahiri PDF

By A. Lahiri

This quantity is an introductory textual content the place the subject material has been provided lucidly in order to aid self examine by means of the novices. New definitions are via 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 options. The ebook starts off with the elemental notions of type, functors and homotopy of continuing mappings together with relative homotopy. basic teams of circles and torus were taken care of besides the basic workforce of masking areas. Simplexes and complexes are awarded intimately and homology theories-simplicial homology and singular homology were thought of besides calculations of a few homology teams. The ebook might be best suited to senior graduate and postgraduate scholars of assorted universities and institutes.

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Example text

We see that F'(x, 0) is a map from C to R. 1 that

If Y is contractible, then every continuous mapping f: X -7 Y is homotopic to a constant. 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.

G = g Proof. Existence off* gives thatf(l) g(l). Since f * is closed, we also have f(O) g(O). 11. So,f- g. 11. This gives that f* null path. g is homotopic to a EXERCISES I. In R2 let A = ((x, y) : x = 0, - 1 Sy S I} and B = ((x, y) : 0 < x S l, y =cos nix}. Show that F = A u B is connected but not path connected. 32 ALGEBRAIC TOPOLOGY 2. In R2 let A = {(x, y) : 0 ~ x ~ l, y = xln, n e N} and B= {

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