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By Martyn R Dixon; Leonid A Kurdachenko; Igor Ya Subbotin

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E:c together 24 ALGEBRA AND NUMBER THEORY: AN INTEGRATED APPROACH (iii) Suppose that a1, a2 E A are such that f(aJ) = j(a2). 1 (f(a2)) so that Since h is injective we deduce that a 1 = a2, which proves that f is injective. 1 is a left inverse of f. (iv) Since h is surjective, for every element c such that f2(a) =c. 1 (f(a)). 1 is also surjective. Furthermore, h o g2 = 8c and therefore so that f o g2 is a right inverse of f 1 • (v) Assertion (iv) implies that the mapping f 1 is surjective and hence, by hypothesis, f 1 is bijective.

16. Construct a bijective mapping from N to Z. 17. Let /! : A -----+ B and h : A -----+ B be mappings. ) and Gr(/2) is a graph of some mapping from A to B if and only if/! 18. Let A and B be finite sets, with injective mappings from A to B. lA I =a, IBI =b. 19. Let A be a set. Prove that A is finite if and only if there exists no bijective mapping from A to a proper subset of A. 20. Let A and B be sets, U <::;A, V <::;B. 3 n f- 1 (V)) = n v. PRODUCTS OF MAPPINGS This section is dedicated to the notion of the product of two mappings.

Put h(b) = ab. Then foh(b) = f(h(b)) = f(ab) = b = es(b), so that f o h = e 8 . This means that h is a right inverse of f. 3 implies that the mapping f must be surjective. SETS 23 The following theorem summarizes some of the main properties of left and right inverses. 6. Theorem. Let f : A --+ B, let /I of. I are injective, then h is also injective. I are surjective, then h is also surjective. I respectively, then g o gi is a right inverse to h· (iii) If the mapping his injective, then f is also injective.

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