The main idea of injective is that f:A-->f(A) be bijective (that is, have an inverse (also a function) f^{-1}:f(A)-->A).

So f is injective if and only if, given b in f(A), there is only ONE a in A with f(a) = b (note that this means there is at LEAST one, because I said "there is", so what I actually mean is there is EXACTLY one).

The "squaring function" on the reals f:R-->R given by f(x) = x^{2} fails this test, for example given the real number 4, we have that the set of pre-images under f for 4 contains TWO numbers: {-2,2}, so the pre-image of 4 is not unique.

The definition of function requires IMAGES, not pre-images, to be unique. This does not precludes the unique image of a number under a function having other pre-images, as the squaring function shows.

To see that this is the same as the classical definition:

f is injective iff: f(a_{1}) = f(a_{2}) implies a_{1} = a_{2},

suppose f(a_{1}) = f(a_{2}) = b. By the first definition I gave, this means there is precisely one a in A, with f(a) = b. Thus we must have a_{1} = a, and similarly a_{2} = a, thus a_{1} = a_{2}.

On the other hand, given the second definition of injective, we can show it implies my first definition:

Suppose whenever f(a_{1}) = f(a_{2}), we have a_{1} = a_{2}. Let b be any element of f(A), so that we know SOME element of a exists with f(a) = b.

Let f^{-1}(b) be the set in A: {x in A: f(x) = b}. Clearly we have {a} is a subset of f^{-1}(b). Now let x be any arbitrary element of f^{-1}(b). By the very construction of the pre-image set, we have f(x) = b. Therefore, we have:

f(x) = f(a), and by the second definition of injective, we have x = a. This shows that f^{-1}(b) is contained in {a}, so f^{-1}(b) = {a}, and we see the pre-image of b is indeed unique.

One final note: logically, any implication is equivalent to its contrapositive. Therefore, the statement:

f(a_{1}) = f(a_{2}) implies a_{1} = a_{2} is often written as its contrapositive:

a_{1} ≠ a_{2} implies f(a_{1}) ≠ f(a_{2}).