# Injective and surjective

1. Feb 2, 2009

### tomboi03

Let f: A-> B
(which i know means.... f is a function from A to B which also means.... that A is the domain and B is the range or image)

Let A0$$\subset$$A and B0$$\subset$$B

a. show that A0$$\subset$$f-1(f(A0)) and the equality hold if f is injective.
b. show that f-1(f(B0)) and B0 and the equality hold if f is surjective

thanks

2. Feb 2, 2009

It follows from the definitions.

Given x in A0, show that it is in f-1(f(A0)). What is f-1(f(A0))? It's the set of all things that map into f(A0). Surely, x is in that set.

To show equality when f is injective, show the reverse inclusion; i.e. show that f-1(f(A0)) is contained in A0. So let x be in f-1(f(A0)), which means that f(x) is an element of f(A0). That means that f(x) = f(x') for some x' in A0; by injectivity, x = x', so x is in A0.

The second part is similar. You really should make a serious attempt at these problems (this and the other two you posted here); they're quite easy. Just make sure you know what the definitions are.

One more thing: f: A -> B does not mean that B is the image of f. It's ok to use words like codomain or target for B, but it's not the image unless f is surjective. Remember, the image of f is the set f(A) = {f(x) | x in A}, which is always a subset of B, but not necessarily all of B. The range of f may refer to either the entire set B or the image of f; it depends on the author. (I personally highly prefer range to mean image, but I typically avoid that term anyway.)

Last edited: Feb 2, 2009