Prove that $f(f^{-1}(V))=f^{-1}(f(V))$ 
Let $f: X\to Y$ be bijective, and $f^{-1}: Y\to X$ be it's inverse. If
  $V\subseteq Y$, show that the forward image of $V$ under $f^{-1}$ is
  the same set as the inverse image of $V$ under $f$.

I have interpreted this as: show that $f(f^{-1}(V))=f^{-1}(f(V))$
I really do not know what to do from here.
 A: No, what they want you to show is that the image of $f^{-1} $:
$$\tag1
(f^{-1})(V)=\{f^{-1}(v):\ v\in V\}
$$
is equal to the preimage of $V $ under $f $:
$$\tag2
f^{-1}(V)=\{x\in X:\ f (x)\in V\}.
$$
The notation is unfortunate in this case, but it is normally used for  $(2) $.
A: Actually, what you have to prove is that
$$f^{-1}(V)=f^{-1}(V)\ .$$
To see why this is not obvious, you have to carefully unpack the meanings of all terms, noting in particular that both $f$ and $f^{-1}$ actually have multiple meanings.
First, the inverse image of $V$ under $f$ is by definition
$$A=\{\,x\in X\ \mid\ f(x)\in V\,\}\ .$$
Second, what is the forward image of $V$ under $f^{-1}$?  We note that since $f$ is a bijection, $f^{-1}$ is a function from $Y$ to $X$ defined by
$$f^{-1}(y)=x\quad\hbox{if and only if}\quad f(x)=y\ .$$
The forward image of $V$ under this function is by definition
$$B=\{\,f^{-1}(v)\ \mid\ v\in V\,\}\ .$$
So, you have to prove that $A=B$, and now I hope you understand why putting it as "$f^{-1}(V)=f^{-1}(V)$" is so confusing!!
Suppose that $x\in A$.  Then
$$\eqalign{f(x)=v\quad \hbox{for some $v\in V$}\quad
  &\Rightarrow\quad x=f^{-1}(v)\quad \hbox{for some $v\in V$}\cr
  &\Rightarrow\quad x\in B\ ,\cr}$$
so $A\subseteq B$.  You also need to show $B\subseteq A$, now that you understand the problem better please try this for yourself.
