If $f :\colon A \rightarrow B$, then $f^{-1}(B) = A$ Let $f\colon A \rightarrow B$. Prove the following statement: $f^{-1}(B) = A$.
My attempts: 
$f^{-1}(B) = \{a\in A\mid f(a)\in B\}$ is a subset of $A$. 
But I can't prove the other way, i.e. $A$ is the subset of $f^{-1}(B)$. I have no idea about it.
 A: Pick $a\in A$. $f(a)\in B$. By definition of $f^{-1}(B)$, $a\in f^{-1}(B)$?
A: Hint: for what elements $a\in A$ can $f(a)\in B$ (that is, $a\in f^{-1}(B)$) fail?
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Here is the proof from a comment of yours, but now reasoning at the logic level.
For every $\;a\;$,
$$\calc
a \in f^{-1}[B]
\calcop{\equiv}{definition of $\;\cdot[\cdot]\;$}
\langle \exists b : f^{-1}(b) = a : b \in B \rangle
\calcop{\equiv}{definition of $\;\cdot^{-1}\;$}
\langle \exists b : f(a) = b : b \in B \rangle
\calcop{\equiv}{logic: one-point rule}
f(a) \in B
\calcop{\equiv}{using the fact that$\;f : A \to B\;$}
a \in A
\endcalc$$
By set extensionality, $\;f^{-1}[B] = A\;$.
Note how this proof separately uses the properties of $\;\cdot[\cdot]\;$ and $\;\cdot^{-1}\;$ in the first three steps.  Usually these would be merged as one step, with a comment like $"\text{definition of $\;\cdot^{-1}[\cdot]\;$}"$.
