What is $f(f^{-1}(A))$? Suppose that $f : E \rightarrow F$.
What is $f(f^{-1}(A))$? Is it always $A$? $f^{-1}$ is the inverse function.
This is not a homework, I'm confused by this statement.
 A: $f[f^{-1}[A]] = \{ f(x): x \in f^{-1}[A] \} = \{f(x): x \in E \text{ such that } f(x) \in A\}$, as the definition of $f^{-1}[A]$ is all $x \in E$ such that $f(x) \in A$. 
So it's all points of $A$ that are actually reached by values of $f$, so $f[E] \cap A$.
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$(This is essentially the same answer as the one by Henno Brandsma, but a bit more expanded in notation I like better, taking smaller steps.)
Which elements $\;y\;$ are in $\;f[f^{-1}[A]]\;$?  Let's expand the definitions.  (Implicitly, we let $\;x \in E\;$ and $\;y \in F\;$.)
$$\calc
y \in f[f^{-1}[A]]
\calcop{\equiv}{definition of $\;\cdot[\cdot]\;$}
\langle \exists x : x \in f^{-1}[A] : f(x) = y \rangle
\calcop{\equiv}{basic property of $\;\cdot^{-1}[\cdot]\;$}
\langle \exists x : f(x) \in A : f(x) = y \rangle
\calcop{\equiv}{logic: substitute for $\;y\;$ in left hand part from right hand part}
\langle \exists x : y \in A : f(x) = y \rangle
\calcop{\equiv}{logic: extract part not using $\;y\;$ out of $\;\exists\;$}
y \in A \;\land\; \langle \exists x :: f(x) = y \rangle
\calcop{\equiv}{make implicit range explicit}
y \in A \;\land\; \langle \exists x : x \in E : f(x) = y \rangle
\calcop{\equiv}{definition of  $\;\cdot[\cdot]\;$}
y \in A \;\land\; y \in f[E]
\calcop{\equiv}{definition of $\;\cap\;$}
y \in A \cap f[E]
\endcalc$$
By set extensionality, $\;f[f^{-1}[A]] \;=\; A \cap f[E]\;$.
