I have no idea how to prove the following, any help on how to resolve this would be appreciated:
Let $\displaystyle f$ be a function from X onto Y and let $\displaystyle B \subset Y$. Show that $\displaystyle f(f^{-1}(B)) = B$.
You need to show two things.
1)If $\displaystyle x\in f(f^{-1}(B))$ then $\displaystyle x\in B$.
2)If $\displaystyle x\in B$ then $\displaystyle x\in f(f^{-1}(B))$.
Together with these steps it shall prove $\displaystyle f(f^{-1}(B)) = B$.
I start you off.
If $\displaystyle x\in f(f^{-1}(B))$ then $\displaystyle x = f(y)$ for some $\displaystyle y\in f^{-1}(B)$.
But $\displaystyle y \in f^{-1}(B)$ means $\displaystyle f(y) \in B$, by definition, thus $\displaystyle x\in B$.
Now you go the other way.
By definition $\displaystyle f^{-1}(B)$ ={ x:xεX & f(x)εB }.....................................1
So we see that $\displaystyle f^{-1}(B) $ IS A subset of X. Now since for any subset ,lets say A, of X we have by definition f(A) ={ f(x) : xεA } then $\displaystyle f( f^{-1}(B))$ ={ f(x): xε$\displaystyle f^{-1}(B)$ }................................................. ...........2
So we see that $\displaystyle f( f^{-1}(B))$ is a set of the images of x f(x) , such that xε$\displaystyle f^{-1}(B)$.
The question now is are these f(x)'s inside B??
Lets see:
Let .........................f(x)ε $\displaystyle f( f^{-1}(B))$ then from 2 we infer f(x)=f(x) & xε$\displaystyle f^{-1}(B)$ .
But from 1 if xε $\displaystyle f^{-1}(B)$ we conclude xεX & f(x)εB.
Hence f(x) is inside B.AND $\displaystyle f( f^{-1}(B))$ is a subset of B
Now for the converse let f(x)εB. Since f is onto there exists an xεX such that f(x)εB AND so by 1 xε$\displaystyle f^{-1}(B)$........................e.t.c,..........e.t.c
For the other way around, does this work?
Say $\displaystyle x \in B \subset Y$
Then there exists an $\displaystyle y \in f^{-1}(B) \subset X$ such that $\displaystyle f(y)=x$, by definition of a surjective function
But then: $\displaystyle f(y) \in f( f^{-1}(B))$ and since $\displaystyle x=f(y)$, we have $\displaystyle x \in f( f^{-1}(B))$ as desired.