Yahoo Answers: Answers and Comments for Sets and Maps, I need some advice on the properties of these maps? [Mathematics]
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From Anonymous
enUS
Sat, 22 Jan 2011 11:46:17 +0000
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Yahoo Answers: Answers and Comments for Sets and Maps, I need some advice on the properties of these maps? [Mathematics]
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From BigBrainOnBrad: 1) you are correct. I'll denote the map wi...
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Sun, 23 Jan 2011 00:03:00 +0000
1) you are correct. I'll denote the map with f.
injectivenes: for any x different from y ==> f(x) different from f(y). This is obvious.
surjectivenes: you would have to find that for any natural y there exists at least a natural x such that y=f(x). So if you find one counterexample the function is not surjective. y=8 let's see what natural number x has the property that x^6 = 8 ==> x = 8^(1/6) = (2^3)^(1/6) = 2^[3*(1/6)] = 2^(1/2) = sqrt(2). Any respectable math student knows that proof that sqrt(2) is not rational (yet alone natural!). So f is not surjective. ==> f is not bijective
2) A map is a relation that associates each element of the domain to one element from the range.
Imagine that the elements of M are k boxes. In each box you can put any one of the l elements of N. Multiply the possibilities and you get l*l*l*...*l*l > that is k times l ==> l^k functions.
I'll give you an example:
f:{a,b,c}>{1,2}
1) f(a)=f(b)=f(c)=1
2) f(a)=f(b)=1 , f(c)=2
3) f(b)=f(c)=1 , f(a)=2
4) f(c)=f(a)=1 , f(b)=2
5) f(a)=f(b)=2 , f(c)=1
6) f(b)=f(c)=2 , f(a)=1
7) f(c)=f(a)=2 , f(b)=1
8) f(a)=f(b)=f(c)=2
you got 2^3 = 8 functions.
3) It's a interval's game here:
f:R>R , f(x)=x^2 is not surjective because 1 = x^2 has no real solution
g:R>R , g(x)=x is not surjective because 100 = x has no real solution
BUT, when we define them like this: f,g:R>[0, infinity) they are surjective because all values in the range have at least one real solution
So,
phi:R>R , phi(x) = x^2 is not surjective
psi:R>[0; infinity) , psi(x) = x
(psi_of_phi):R>[0; infinity) , (psi_of_phi)(x) = psi(phi(x)) = x^2 which is surjective as noticed above.
So, here are your examples.
Best regards!

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Sat, 29 Oct 2016 03:22:23 +0000
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From Jim R: 1. You're right.
2. Since M has k eleme...
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Sat, 22 Jan 2011 22:33:28 +0000
1. You're right.
2. Since M has k elements, let's name the elements: M = {x1, ..., xk}. A map, or function, from M to N consists of k ordered pairs, each one looking like {xi, y}, where 1 <= i <= k and y is some member of N. There are l (lowercase L) choices for y in each of those ordered pairs, so the total number of ways you can choose the y's is l times l ... times l (k factors) or l^k. Definitely not an infinite number, to answer your followup question.
3. Let's make psi of phi bijective. (You said surjective the first time, but if it's bijective it'll certainly be surjective.) If M, N, and P can be any sets, this is actually pretty easy. For example, let M and P be the nonnegative real numbers, and let N be all the real numbers. Let phi be the identity function (in other words, phi(x) = x for all x), and let psi be the absolute value function. Then phi is not surjective, because its range doesn't contain any negative numbers, but psi of phi of any nonnegative real number x is x, so psi of phi is the identity function from M to P.
Notice that psi isn't injective, let alone bijective, on the set of all real numbers (for example, psi(1) = psi(1) = 1); however, its restriction to the nonnegative real numbers (which is what psi of phi amounts to) is injective and in fact bijective.