# Show that if f: A→B is surjective and and H is a subset of B, then f([

1. Aug 20, 2013

### Scienticious

[f]^{}[/2]1. The problem statement, all variables and given/known data

Show that if f: A→B is surjective and and H is a subset of B, then f(f^(-1)(H)) = H.

2. Relevant equations

3. The attempt at a solution

Let y be an element of f(f^(-1)(H)).
Since f is surjective, there exists an element x in f^(-1)(H) such that f(x) = y.
But x in f^(-1)(H) implies that f(x) is in H, by definition of inverse functions.
Therefore f(f^(-1)(H)) is a subset of H.

Conversely, for y in H y must also be in f(f^(-1)(H)) since the latter expression is equivalent to H by definition
of inverse functions.

Since f(f^(-1)(H)) is a subset of H and vice versa, f(f^(-1)(H)) = H.

Last edited: Aug 20, 2013
2. Aug 20, 2013

### Office_Shredder

Staff Emeritus
This part is good

This part less so. What if f(x) = 0 for all x a real number, and H is the interval [0,1]? What is f(f-1(H)) in this case?

You have to use the fact that f is surjective.

3. Aug 20, 2013

### Scienticious

O I understand now I think, thank you :3

f is surjective, so for any y in H we have f(x) = y.
since f^(-1)(H) is in the domain of f by definition of inverse functions,
y = f(f^(-1)(y)) for any y in H and thus H is a subset of f(f^(-1)(H)).

This line of reasoning checks out, right?

4. Aug 20, 2013

### verty

Consider using this type of argument: $A \subseteq B \land A \supseteq B → A = B$.

5. Aug 20, 2013

### vela

Staff Emeritus
You need to be a bit more careful. Consider the case of f: ℝ→{0,∞} where f(x)=x2. The mapping $f^{-1}$ isn't a function, which you seem to be assuming. What would $f^{-1}(4)$ be equal to, for instance — 2 or -2?