What is the relationship between f-1(f(A0)) and A0 in terms of injectivity?

[tomboi03]

I'm not sure how i would go about this problem...

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

 

[adriank]

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.)