By John Bonnycastle

**Read or Download An Introduction To Mensuration And Practical Geometry; With Notes, Containing The Reason Of Every Rule PDF**

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**Extra info for An Introduction To Mensuration And Practical Geometry; With Notes, Containing The Reason Of Every Rule**

**Example text**

7rn(B) or, by the Hurewicz theorem, f* : G -. 7rn(B). The transformation -. f* yields a natural homomorphism q: 7rn(G, B) -. Hom(G, 7rn(B)). n f f of K I (G, n) - K ' (G, n) v K I (G, n) is homotopy-commutative, for any f But the diagram is clearly commutative for f = Eg, g : Sn - i --+ K'(G, n -1) and every map f: Sn-*K'(G, n) 29 30 HOMOTOPY THEORY AND DUALITY is homotopic to Eg, for some g. Thus tj is a homomorphism. If G is free, , is easily seen to be an isomorphism. Furthermore, we have the following commutative diagram, based on q : G1-.

Y5 HS 1 G6 INDUCED FIBRE AND COFIBRE MAPS 37 the rows are exact and there is commutativity in each square, then the sequence ... - G Y1-+1 1 H+ 1 G2' 1 2) H 2 3 2 0 64 Y4- H4Q G5 4 , .. is exact. Proof. It is trivial to verify that the image is contained in the kernel at each term. We show the opposite inclusion first at H4 +Q G5. We are given x s H4, y s G5 such that /4x+ Y5y =0. Now 0 05y = 5Y5y = so that Y= 04z for some z. ll/4(x+Y4Z) = 1f/4x+Y54)4z = 0. Thus x +Y4z = Y' 3t = Y44) 3 t for some t.

If h : G1-+ G2 is a homomorphism of abelian groups, then, by a standard obstruction argument, there is a map j3 : K(G1, m) -+ K(G2, m) inducing h in the sense that h is the induced homomorphism of mth homotopy groups. Moreover, the homotopy class of fi is determined by h. Let us factor fi through the mapping track and replace fi by the associated fibre map. If we suppose that h is an epimorphism then the fibre is a K(Go, m) where Go is the kernel of h. This is seen by taking the homotopy sequence of the fibre map.

### An Introduction To Mensuration And Practical Geometry; With Notes, Containing The Reason Of Every Rule by John Bonnycastle

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