Solving Mappings Problem: Onto, One-to-One and Co-domain

[jackiemoon]

This is very basic but I'm having doubts about my understanding of mappings, onto, one-to-one etc and the textbook I'm using isn't very clear.

1. Let α,β and γ be mappings from Z to Z defined by

α(n) = 3n
β(n) = 4n²+1
γ(n) = 2 + cos (nπ/2) ...this is (n(pi)/2


Which of the 3 mappings are onto or one-to-one, and determine which subset is the co-domain of the mapping γ?



2.



3. My understanding is, taking α(n) for example, 0 maps to 0, 1 maps to 3, -1 maps to -3 etc, means that the mapping is one-to-one but not onto, because not all integers in the codomain are mapped to e.g. -2, 2. If we are told the mappings are from Z to Z, does that mean than all integers are included in each set?I'm probably not being very clear about this...apologies but I have scoured the net for a clear definition or suitable example but can't find anything. Any help would be appreciated.

 

[CompuChip]

So indeed a is one-to-one (or injective), because it maps different numbers to different numbers (that is, if a(x) = a(y) then x = y, which is the definition of injectivity). Not all elements of the codomain have a pre-image, so it is not onto (or surjective.)
If we say the mapping is from A to B then we mean, that there is precisely one image for every element of A (so a function cannot map one element to two different ones at the same time). Usually, we mean that B is the codomain, such that f(a) lies in B for every a in A. We then distinguish the range, which is a subset of the codomain, as all those points of the codomain which are actually mapped to by some element of the domain. Now if every element of the range is mapped to precisely once (in other words: every element of the codomain is mapped to no more than once) the function is called one-to-one or injective. If the codomain is equal to the range (in other words, every element of the codomain is mapped to at least once) the function is called onto or surjective. Functions which are both are also called bijections. In a sense, they say that two sets are the same because we can go from one to the other by the bijection in a unique, one-to-one, way.

To give a quick example:
f(x) = x for all x in R and f(0) = 6 is not a function, because it tries to map 0 to two different values;
f: R -> R, f(x) = x^2 is a function, though it is not injective (-2 and 2 are both mapped to 4) and not surjective (-1 does not have a pre-image);
f: R⁺ -> R, f(x) = x^2, where R⁺ denotes the set of all non-negative reals, is injective but not surjective;
f: R -> R⁺ is not injective, but surjective;
finally,
f: R⁺ -> R⁺ is bijective (that is, injective and surjective).