1. ## surjective proof

I have no idea how to prove the following, any help on how to resolve this would be appreciated:

Let $f$ be a function from X onto Y and let $B \subset Y$. Show that $f(f^{-1}(B)) = B$.

2. Originally Posted by lllll
I have no idea how to prove the following, any help on how to resolve this would be appreciated:

Let $f$ be a function from X onto Y and let $B \subset Y$. Show that $f(f^{-1}(B)) = B$.
You need to show two things.
1)If $x\in f(f^{-1}(B))$ then $x\in B$.
2)If $x\in B$ then $x\in f(f^{-1}(B))$.
Together with these steps it shall prove $f(f^{-1}(B)) = B$.

I start you off.
If $x\in f(f^{-1}(B))$ then $x = f(y)$ for some $y\in f^{-1}(B)$.
But $y \in f^{-1}(B)$ means $f(y) \in B$, by definition, thus $x\in B$.

Now you go the other way.

3. Originally Posted by lllll
I have no idea how to prove the following, any help on how to resolve this would be appreciated:

Let $f$ be a function from X onto Y and let $B \subset Y$. Show that $f(f^{-1}(B)) = B$.
By definition $f^{-1}(B)$ ={ x:xεX & f(x)εB }.....................................1

So we see that $f^{-1}(B)$ 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 $f( f^{-1}(B))$ ={ f(x): xε $f^{-1}(B)$ }................................................. ...........2

So we see that $f( f^{-1}(B))$ is a set of the images of x f(x) , such that xε $f^{-1}(B)$.

The question now is are these f(x)'s inside B??

Lets see:

Let .........................f(x)ε $f( f^{-1}(B))$ then from 2 we infer f(x)=f(x) & xε $f^{-1}(B)$ .

But from 1 if xε $f^{-1}(B)$ we conclude xεX & f(x)εB.

Hence f(x) is inside B.AND $f( f^{-1}(B))$ 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 $f^{-1}(B)$........................e.t.c,..........e.t.c

4. Originally Posted by ThePerfectHacker
You need to show two things.
1)If $x\in f(f^{-1}(B))$ then $x\in B$.
2)If $x\in B$ then $x\in f(f^{-1}(B))$.
Together with these steps it shall prove $f(f^{-1}(B)) = B$.

I start you off.
If $x\in f(f^{-1}(B))$ then $x = f(y)$ for some $y\in f^{-1}(B)$.
But $y \in f^{-1}(B)$ means $f(y) \in B$, by definition, thus $x\in B$.

Now you go the other way.
For the other way around, does this work?

Say $x \in B \subset Y$
Then there exists an $y \in f^{-1}(B) \subset X$ such that $f(y)=x$, by definition of a surjective function
But then: $f(y) \in f( f^{-1}(B))$ and since $x=f(y)$, we have $x \in f( f^{-1}(B))$ as desired.

5. Originally Posted by Last_Singularity
For the other way around, does this work?

Say $x \in B \subset Y$
Then there exists an $y \in f^{-1}(B) \subset X$ such that $f(y)=x$, by definition of a surjective function
But then: $f(y) \in f( f^{-1}(B))$ and since $x=f(y)$, we have $x \in f( f^{-1}(B))$ as desired.
Yes it works.
It indeed has to be a surjective function;