EDIT: Proving $f^{-1}(f(C))=C$ I need to prove that $f^{-1}(f(C))=C$.
This are the informations. There exists two sets A and B, and function $f(A)\to B$. 
I don't know how to solve this, and I tried to search google, but I didn't find anything useful.
Please help. Thanks!!
EDIT: I forgot to add that function is bijective
 A: Hint: Show that $f^{-1}(f(C)) \subseteq C$ and $f^{-1}(f(C)) \supseteq C$. One of these is true for any $f$, while the other relies on the fact that $f$ is bijective (in fact, you just need $f$ to be injective).
A: Here is a somewhat bigger hint, in a different style than you may be used to.
First, which elements $\;x\;$ does the set $\;f^{-1}[f[C]]\;$ contain?  Let's use the basic properties and calculate:
\begin{align}
& x \in f^{-1}[f[C]] \\
\equiv & \;\;\;\;\;\text{"basic property of $\;\cdot^{-1}[\cdot]\;$: $\;x \in f^{-1}[W] \;\equiv\; f(x) \in W\;$"} \\
& f(x) \in f[C] \\
\equiv & \;\;\;\;\;\text{"basic property of $\;\cdot[\cdot]\;$: $\;y \in f[V] \;\equiv\; \langle \exists x : x \in V : f(x) = y \rangle\;$"} \\
& \langle \exists z : z \in C : f(z) = f(x) \rangle \\
\text{...} & \;\;\;\;\;\text{"..."} \\
\end{align}
Now we want to end this calculation with $\;x \in C\;$ (why?), and there are two ways to continue it:
\begin{align}
& \langle \exists z : z \in C : f(z) = f(x) \rangle \\
\Leftarrow & \;\;\;\;\;\text{"choose a specific $\;z\;$ that gets us near our goal..."} \\
& \text{...} \\
\end{align}
and also
\begin{align}
& \langle \exists z : z \in C : f(z) = f(x) \rangle \\
\Rightarrow & \;\;\;\;\;\text{"assume $\;f\;$ has some specific property to achieve our goal..."} \\
& \text{...} \\
\end{align}
How do you complete the calculations?  What property of $\;f\;$ do you need?  What is your conclusion?
