# Set function proof

1. Oct 13, 2011

### autre

I need to prove f(f^-1(Y'))$\subseteq$Y' for some f: X -> Y and Y' in Y.

So far, I've been able to figure this much out:

Let y$\in$f(f^-1(Y')). Then, f^-1(Y') = x' for some x' in X such that f(x') = y' for some y' in Y'. Then, f(x') = y'. Thus, f(f^-1(Y'))$\subseteq$Y'.

I feel like there's something wrong with my proof. Any ideas on where I went wrong?

2. Oct 13, 2011

### autre

Any ideas?

3. Oct 13, 2011

### micromass

You probably mean Y' subset of Y.

What does that even mean?? Y' is a set, so $f^{-1}(Y^\prime)$ is a set. But x' is an element. So you're saying that a set is equal to an element????

Start with $y\in f(f^{-1}(Y^\prime))$. Write out the definitions. What is the definition for $y\in f(A)$??

4. Oct 13, 2011

### autre

Thanks for the input micromass.

I've revised the proof as thus:

Let y∈f(f^-1(Y')). Since f^-1(Y') = X' s.t. X'⊆X and f(X')=Y'. Since y∈f(X'), y∈Y'. Thus, f(f^-1(Y'))⊆Y'.

I still feel like I'm missing a step or two.

5. Oct 13, 2011

### micromass

Why??

This would imply

$$f{-1}(f(X^\prime))=X^\prime$$

which does not always hold.

6. Oct 13, 2011

### autre

You're right, I can't assume f is bijective. Should I instead say something like "there exists an x∈X s.t. x∈f^-1(Y')?

7. Oct 13, 2011

### Bacle2

Maybe a good comment to make is that, in the nicest-possible case, you have:

f-1of(X)=fof-1X=X.

Nicest possible is, of course, f is 1-1 and onto. Try to see why identity above fails when f is either not 1-1 or not onto.

TMFKAB (The Mathematician* Formerly Known as Bacle)

*In training.

8. Oct 13, 2011

### autre

Not sure how that helps, Bacle2. Basically, if the function isn't surjective there could exist a b in B' such that f^-1(B') doesn't exist, and I'm not sure how to handle this case.

9. Oct 13, 2011

### micromass

Bacle wasn't giving you a hint to solve the problem. He gave you another problem which could be rewarding to look at to expand your knowledge.

To solve your problem. What does it mean that $y\in f(A)$. Apply this to $y\in f(f^{-1}(y))$.

10. Oct 13, 2011

### autre

I think I follow. You mean something like:

Let y∈f(f^-1(Y')). Then, there exists an x in f^-1(Y') s.t. f(x) = y. Since x in f^-1(Y'), f(x) = y for some y in Y'. Thus, f(f^-1(Y'))⊆Y'.

11. Oct 13, 2011

### micromass

Seems ok.

12. Oct 13, 2011

### Bacle2

Yes, Autre, sorry if my post was confusing; just trying to give some insight and some related results, as Micromass said.