# 36c - tion and inverse relations oF Functions The...

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p ( b ) = some a X such that f ( a ) = b . It does not matter which a we choose, but there will be such an a by defi- nition of f [ X ] . We are placing the members of f [ X ] in the pigeonholes X . By the pigeonhole principle, some pigeonhole has at least two occupants. In other words, there is some a X and b, b f [ X ] with p ( b ) = p ( b ) = a . But then f ( a ) = b and f ( a ) = b , which cannot happen as f is a function. P ROPOSITION 4.13 Let A and B be finite sets, and let f : A B . 1. If f is one-to-one, then | A | | B | . 2. If f is onto, then | A | | B | . 3. If f is a bijection, then | A | = | B | . Proof Part (a) is the contrapositive of the pigeonhole principle. For (b), notice that if f is onto then f [ A ] = B , so that in particular | f [ A ] | = | B | . Also | A | | f [ A ] | by proposition 4.12. Therefore | A | | B | as required. Finally, part (c) follows from parts (a) and (b). 4.5 Operations on Functions Since functions are special relations, we can define the identity, composi-
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Unformatted text preview: tion and inverse relations oF Functions. The composition oF two Functions is always a Function. In contrast, we shall see that the inverse relation oF a Function need not necessarily be a Function. D EFINITION 4.14 (C OMPOSITION OF ± UNCTIONS ) Let A,B and C be arbitrary sets, and let f : A → B and g : B → C be arbitrary Functions oF these sets. The composition oF f with g , written g ◦ f : A → C , is a Function defned by g ◦ f ( a ) def = g ( f ( a )) For every element a ∈ A . In Haskell notation, we would write (g.f) a = g (f a) It is easy to check that g ◦ f is indeed a Function. Notice that the co-domain oF f must be the same as the domain oF g For the composition to be well-defned. 36...
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