Demonstrating the image of the inverse image of a subset I need to demonstrate the following:
Let $E, F$ be sets, $Y \subset F$ and $f : E\longrightarrow F$.
Prove that $f(f^{-1}(Y)) = Y \cap f(E)$
I tried do prove that using double inclusion with multiple techniques, such as direct & inverse image properties, or using cardinals, but I'm always stuck at some point.
Some of my research, if that can help you (sorry for the french I hope you still understand) :

(don't pay attention to the last line of the right side, I know it's false)
 A: Okay, let us look at the first part- to prove that f(inv(f(Y))) C (Y int f(E)).
Let x belong to f(inv(f(Y))), then there exists an e in inv(f(Y)) such that x= f(e), but that implies that there exists a t in Y such that e = inv(f(t)), so f(e) = t = x, but since t is in Y, and t=x, x is in Y ... (1)
Now, since x = t = f(e), and e is in inv(f(Y)) C E, e is in E, and x = f(e), so x is in f(E) ... (2)
Thus, x is in Y and f(E), from (1) and (2), so we have x is in (Y int f(E)), so we have f(inv(f(Y))) C (Y int f(E)) ... [A]
Now, we prove that (Y int f(E)) C f(inv(f(Y))). Now, let p be in (Y int f(E)), so there exists an e in E such that p = f(e) and p is in Y, and since p is in Y, inv(f(p)) is in inv(f(Y)), so e is in inv(f(Y)), so e = inv(f(p)), so f(e) = f(inv(f(p))), so f(e) is in f(inv(f(Y))), so p = f(e) is in f(inv(f(Y))). Thus, we have (Y int f(E)) C f(inv(f(Y))) ... [B]
From [A] and [B], we have f(inv(f(Y))) = (Y int f(E)), and hence, we are done.
A: If $Y \subset F$ and $f : E \to F$, 
then $f(f^{-1}(Y)) = Y \cap f(E)$
\begin{align}
   y \in f(f^{-1}(Y))
   &\implies \exists x \in f^{-1}(Y),\, y = f(x)\\
   &\implies (x \in E) \wedge (f(x) \in Y)\\
   &\implies f(x) \in Y \cap f(E)\\
   &\implies y \in Y \cap f(E)\\
\end{align}
So $f(f^{-1}(Y)) \subseteq Y \cap f(E)$
\begin{align}
   y \in Y \cap f(E)
  &\implies (y \in Y) \land (y\in f(E))\\
  &\implies (y \in Y) \land (\exists x \in E,\, y=f(x))\\
  &\implies \exists x \in E,\, x \in f^{-1}(y)\\
  &\implies y \in f(f^{-1}(y))\\
\end{align}
So $Y \cap f(E) \subseteq f(f^{-1}(Y))$.
