# Preimage proof?

1. Sep 16, 2012

### SMA_01

1. The problem statement, all variables and given/known data

Define f:S→T, where B$\subseteq$T. Let f-1(B)={x$\in$S:f(x)$\in$B} be the preimage of B.

Demonstrate that for any such map f, f(f-1(B))=B.

My main question is, would I prove this using set inclusion both ways?

I was going to begin by letting an element be in the preimage of B, and explain what that means, then mapping that element to B. Would this be correct?

I just need a push in the right direction.

Thank you.

2. Sep 16, 2012

### SammyS

Staff Emeritus
Generally, yes. To show set equality you need to show inclusion in both directions.

3. Sep 16, 2012

### micromass

Just to let you know, but that equality is false in general. You need extra assumptions on f.

4. Sep 16, 2012

### SMA_01

SammyS- Thank you

Micromass- Yes, the function needs to be surjective, right? I'm not sure why there were no assumptions...

5. Sep 16, 2012

### SammyS

Staff Emeritus
I'm very reluctant to question you, micromass.

But isn't the extra assumption of f needed for $f^{-1}\left(f(A)\right)=A$ rather than for $f\left(f^{-1}(B)\right)=B\ ?$

6. Sep 16, 2012

### micromass

An extra assumption is needed for both. Consider $f(x)=x^2$ and B=[-1,0]. Then $f^{-1}(B)=\{0\}$ and thus $f(f^{-1}(B))=\{0\}$. The problem is that f is not surjective.

7. Sep 16, 2012

### SammyS

Staff Emeritus
Well, my reluctance was well founded!

Thanks as always.