Let f : S rightarrow T be a function. For Y T let f -1(Y) denote the inverse set

Question

Let f : S rightarrow T be a function. For Y T let f -1(Y) denote the inverse set function (see Naylor and Sell, Ch.2, pg. 27). F-1 (Y) is the largest subset of S which F maps into Y; that is, Prove the following for arbitrary| subsets X of 5 and Y of T: In addition to the function , there is another mapping of subsets into subsets which is naturally associated with f. It maps P(Y) into P(X). For the moment we will denote it -1. We define -1 as follows: where C ? P(Y). That is, -1(C) consists of all pre-image points of points in C. The mapping -1 is referred to as the inverse set function or inverse image mapping and the set -1(C) is said to be the inverse image of C. Note that

Explanation / Answer

Theorem: Do you mean that f is invertible and g is the inverse of f ? In this case your proof for forward images obviously goes through. Proof: ------- But in fact, given a map f:S?T , the map that assigns to every set A?T the set f -1 [A] is actually a Boolean homomorphism. I.e., it not only preserves unions (as taking forward images does) but also intersections and complements. The arguments for this are quite simple. Let me give you the details for unions. Let A,B?T . If a?f -1 [A]?f -1 [B] , then there is b?A or b?B such that f(a)=b . Hence a?f -1 [A?B] . If a?f -1 [A?B] then there is b?A?B such that f(a)=b . So there is b?A with f(a)=b or there is b?B such that f(a)=b . In other words, a?f -1 [A]?f -1 [B] . This shows f -1 [A?B]=f -1 [A]?f -1 [B] . Observe that if f happens to be invertible and g is the inverse of f , then for all A?T , f -1 [A]=g[A] . Oh, and don't be confused by my use of square brackets. I use these when I am taking images or preimages of sets rather than points, to avoid confusion. Also note that forward images in general don't preserve intersections. Conclusion: so f is invertible set of g

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