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2. Inverse functions Inverse relation, Function is a relation Is the function’s inverse relation a function? No Example: A={1,2,3},B={a,b}, f:A→B, f={(1,a),(2,b),(3,b)} is a function, but inverse relation f -1 ={(a,1),(b,2),(b,3)} is not a function.
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Theorem 3.7: Let f be a function from A to B, then inverse relation f -1 is a function if only if f is one-to-one correspondence. Proof: (1)If f –1 is a function, then f is one-to-one correspondence. (i)f is onto. For any b B , there exists a A such that f (a)=?b (ii)f is one to one. If there exist a 1,a 2 A such that f(a 1 )=f(a 2 )=b B, then a 1 ?=a 2 (2)If f is one-to-one correspondence , then f –1 is a function f -1 is a function, for any b B , there exists one and only a A so that (b,a) f -1. For any b B, there exists a A such that (b,a) ?f -1. For b B , If there exist a 1,a 2 A such that (b,a 1 ) f -1 and (b,a 2 ) f -1,then a 1 ?=a 2
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Definition 3.5: Let f be one-to-one correspondence between A and B. We say that inverse relation f -1 is the inverse function of f. We denoted f -1 : B→A. And if f (a)=b then f -1 (b)=a. Theorem 3.8: Let f be one-to-one correspondence between A and B. Then the inverse function f -1 is also one-to-one correspondence. Proof: (1) f –1 is onto (f –1 is a function from B to A For any a A,there exists b B such that f -1 (b)=a) (2)f –1 is one to one For any b 1,b 2 B, if b 1 b 2 then f -1 (b 1 ) f -1 (b 2 ). If f:A→B is one-to-one correspondence, then f -1 : B→A is also one-to-one correspondence. The function f is called invertible.
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Theorem 3.9: Let f be one-to-one correspondence between A and B. Then (1)(f -1 ) -1 = f (2)f -1 f=I A (3)f f -1 =I B Proof: (1) (2)
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Let f:A→B and g:B→A , Is g the inverse function of f ? f g?=I B and g f ?=I A Theorem 3.10 : Let g be one-to-one correspondence between A and B, and f be one-to-one correspondence between B and C. Then (f g) -1 = g -1 f -1 Proof: By Theorem 3.6, f g is one-to-one correspondence from A to C Similarly, By theorem 3.7, g -1 is a function from B to A, and f –1 is a function from C to B.
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Theorem 3.11: Let A and B be two finite set with |A|=|B|, and let f be a function from A to B. Then (1)If f is one to one, then f is onto. (2) If f is onto, then f is one to one. The prove are left your exercises.
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Exercise: P176 21,22 Prove T 3.11 Cardinality Paradox Pigeonhole principle P88 3.3
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