The, I suppose "technical", name for this mapping is modulo 2 - if is even then for some and so has remainder 0 when divided by 2. Similarly, if is odd then for some and so has remainder 1 when divided by 2.
Let us look at an arbitrary number, . As is an odd number, f will be as the remainder when you divide the number by is just . . Does that make sense?
Now, note that . This means that the mapping takes numbers from the natural numbers and maps them to the set . Every number is mapped to a single number - this is what a mapping is! Of course, more than one number is mapped to 1, and more than one number is mapped to 0. We call - the stuff that we are mappiing from - the domain, and we call - the stuff we are mapping to - the range (also known as the codomain).
So, a function is injective (in shorthand, 1-1) if implies that . That is to say, every element in the range is mapped to by one and only one element. In this example but and so this mapping is not injective.
Can you see how this relates to the concept of injective (1-1) mappings? I mean, just look at the definition I gave you, above!
(Note that 1 is coprime to every number as no number other than 1 divides 1.)
So, . Now, is coprime to , is coprime, is not, is, is not, is not (note the ), is, is, is not, is not, is, is not, is, and is. Counting all the is-es, we see that there are numbers less than that are coprime to . Thus, .
To find all the numbers quickly, what you need to do is write all the numbers less than your number down - . Then, write down each prime number that divides your number (no repeats though). Then, score off , the first prime in your list, and then 2 , etc, then go to the second prime in your list and do the same, and so on. Then score off 1. All the numbers that are left are the numbers that are coprime! Simples!
So, that is what the mapping does, and the question is asking you to see where and are mapped to.