Proof of Set Function f(f^-1(Y'))\subseteqY

[autre]

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

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

Let y\inf(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'))\subseteqY'.

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

 

[autre]

Any ideas?

 

[micromass]

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

You probably mean Y' subset of Y.

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

Let y\inf(f^-1(Y')). Then, f^-1(Y') = x'

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)??

 

[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.

 

[micromass]

Since f^-1(Y') = X' s.t. X'⊆X and f(X')=Y'.

Why??

This would imply

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

which does not always hold.

 

[autre]

Why??

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')?

 

[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.

 

[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.

 

[micromass]

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.

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)).

 

[autre]

What does it mean that y∈f(A). Apply this to y∈f(f−1(y)).

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'.

 

[micromass]

Seems ok.

 

[Bacle2]

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