Functions and InversesDate: 05/29/2002 at 01:42:57 From: Marianne Richa Subject: Functions (Surjection, Bijection) Hi Dr. Math, I was doing the following exercise and I couldn't find when both are injective and surjective: Let A, B, and C be three sets, and f:A->B, g:B->C are two functions. Define A, B, C, f and g such that (g o f) is both injective and surjective, but f is not surjective and g is not injective. Thanks for your help. Best Regards Marianne Date: 05/29/2002 at 12:53:52 From: Doctor Peterson Subject: Re: Functions (Surjection, Bijection) Hi, Marianne. Let's try to picture it. Since f is not surjective, the image of A (range of f) must be only part of B: +---------------+ | B | +-----+ | +-----+ | | | f | | | | | A |--------------->| A' | | | | | | | | +-----+ | +-----+ | | | +---------------+ But when we follow that with g, the composition must be both injective and surjective, so C must be "the same size as" A, and must be the image of A': +---------------+ | B | +-----+ | +-----+ | +-----+ | | f | | | | g | | | A |--------------->| A' |----------------->| C | | | | | | | | | +-----+ | +-----+ | +-----+ | | +---------------+ Now g is defined on all of B, so clearly it must not be injective; it must take the rest of B into C as well. There are many ways to do this. One easy way is to take A, B, and C as the integers, and take f as f(x) = 2x This is not surjective, but is injective; its image is the even integers. Now can you find a g that will take these to all integers? - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 05/29/2002 at 14:23:51 From: Marianne Richa Subject: (Continued) Hi, I understood all that you wrote, but still I couldn't define g(x). Could it be 3x? Thank you. Date: 05/29/2002 at 14:41:59 From: Doctor Peterson Subject: Re: (Continued) Hi, Marianne. You don't sound very sure ... so you probably aren't right! Did you try finding the composite (f o g) and seeing if it fits the requirements; or check whether your g is surjective? You want a function g that is sort of an inverse of f; for every integer, there should be some even number that g will take there. Since we made f double each integer, what if g divides by 2? The problem is, the odd integers won't go to integers then, so you'll need a different definition for the function in that case. If you think about it, you'll realize that it really doesn't matter what you choose to do with the odd numbers (which are B-A' in my picture). You might like to try a different solution. Suppose we just made A = {1,2,3} and B = {4,5,6,7} and C = {8,9,10}. Can you follow my picture to define f and g by just naming where each function takes each element of its domain? The concepts of injective and surjective functions become almost trivial if you learn to picture them right. This page may help a little: http://mathforum.org/library/drmath/view/52454.html You can find more help by searching our site for those words. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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