Modern algebra : an introduction by John R Durbin

By John R Durbin

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D) A sufficient condition for a to be invertible is that it be one-to-one. 12. For a a mapping, decide whether each of the following is true or false. (a) a is invertible only if a is onto. (b) a is invertible if a is onto. (c) A sufficient condition for a to be onto is that it be invertible. (d) A necessary condition for a to be onto is that it be invertible. 13. Consider f and g, mappings from JR to JR, defined by f(x) equal to g 0 f? = sin x and g(x) = 2x. 14. Which of the functions sine, cosine, and tangent, as mappings from JR to JR, are invertible?

The numbers under k and a(k) in the two-row form of fJ are fJ(k) and (fJ 0 a)(k), respectively. But, since fJ- 1 0 fJ = l (the identity mapping), fJ 0 a = fJ 0 a 0 (fJ- 1 0 fJ) = (fJ 0 a 0 fJ-I) 0 fJ, so the number following fJ(k) in the cyclic decomposition of fJ 0 a 0 fJ- 1 is (fJ 0 a)(k). 1. Assume a = C 2 3 4 3 ~)and~=C Compute each of the following. (b) a o~ (a) ~ 0 a (f) a-I 0 (e) ~-I 0 a-I 3 3 1 4 ~). (d) ~-I (h) (a 0 ~)-I (c) a-I (g) (~o a)-I ~-I . 2. 1 usmg a = 2 2 3 4 ~) and~ = (: 2 3 3 ~).

Define 0: o:(x) = 2, o:(y) = I, and o:(z) = 3, : S --* T by as shown in the diagram below. Also, define fJ : T --* U by fJ(l) Then = b, fJ(2) = c, and fJ(3) = a. = fJ(o:(x» = fJ(2) = c = fJ(o:(y» = fJ(I) = b (fJ oo:)(z) = fJ(o:(z» = fJ(3) = a. 2. Let 0: and fJ denote the mappings, each with the set of real numbers as both domain and codomain, defined by o:(x) = x 2 +2 and fJ(x) = x - 1. Then (0: 0 fJ)(x) = o:(fJ(x» = o:(x - 1) = (x - 1)2 = x2 2x - +2 + 3, while (fJ 0 o:)(x) = fJ(o:(x» + 2) = (x + 2) = x 2 + 1.

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