Injective, Surjective, and Bijective

Given a mapping (function) f from A to f(A):

Definition: f is injective if

1) x1=x2 -> f(x1)=f(x2) Ex: sqrt(4)=+2, sqrt(4)=-2

2) X1≠x2 -> f(x1)≠f(x2) Ex: 2^{2}=(4), (-2)^{2}=4

1) and 2) imply the alternate definition:

3) f(x1)=f(x2) -> x1=x2

4) f(x1)≠f(x2) -> x1≠x2

1) & 4) are equivalent.

2) & 3) are equivalent.

Any of the combinations (tests) 1),2); 1),3); 4),2); 4),3) establish an injection.

Surjective is relative:

If B=f(A), f:A->B is surjective. (if f is also injective, called bijective, or 1-1 onto,)

If B=f(A) is a subset of C, f:A->C is not surjective. (if f is injective, called 1-1 into,)

Re: Injective, Surjective, and Bijective

Quote:

Originally Posted by

**Hartlw** Given a mapping (function) f from A to f(A):

Definition: f is injective if

1) x1=x2 -> f(x1)=f(x2) Ex: sqrt(4)=+2, sqrt(4)=-2

No, that is the definition of "function" itself. Further, sqrt(x), as a function, is defined as " is the **positive** number, a, such that . is NOT equal to -4 by the standard definition.

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2) X1≠x2 -> f(x1)≠f(x2) Ex: 2^{2}=(4), (-2)^{2}=4

Now THIS is the definition of "injective". But your "example" is of a function that is NOT injective.

Quote:

1) and 2) imply the alternate definition:

3) f(x1)=f(x2) -> x1=x2

4) f(x1)≠f(x2) -> x1≠x2

(3) is good. (4) is just the definition of "function".

Quote:

1) & 4) are equivalent.

2) & 3) are equivalent.

Any of the combinations (tests) 1),2); 1),3); 4),2); 4),3) establish an injection.

Again, (1) and (4) are necessary that the relation be a **function** and "injective" and "surjective" are only defined for functions.

Quote:

Surjective is relative:

If B=f(A), f:A->B is surjective. (if f is also injective, called bijective, or 1-1 onto,)

Equivalently, f:A->B is "surjective" if and only if, for any b in B, there exist a in A such that f(a)= b.

Quote:

If B=f(A) is a subset of C, f:A->C is not surjective. (if f is injective, called 1-1 into,)

Re: Injective, Surjective, and Bijective

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}).

Re: Injective, Surjective, and Bijective

HallsofIvy:

1) and 2) are both necessary for f to be an injection. The examples make this clear.

If you are saying sqrt(4) = +2, that is obviously true by convention to make sqrt single-valued, but you missed the point.

Sorry you, emakarov, and Deveno didn’t understand my post, which was concise, clear, correct, and comprehesive.

Re: Injective, Surjective, and Bijective

If three different people did not understand your post then possibly it was NOT as "concise, clear, correct, and comprehensive" as you think! I would not think that defining a property and then giving, as an "example", something that does **not** have that property is at all "clear".

Re: Injective, Surjective, and Bijective

sqrt(x), without + convention, is not injective becaues it doesn’t satisfy 1).

x^{2}, (without domain restriction), is not injective because it doesn’t satisfy 2).

I’m trying to avoid verbose elucidation which becomes tedious, time wasting, difficult to follow, impossible to remember, and obscures the main point.