Prove that if $f:X\to Y$ is bijective, then $(f^{-1})^{(-1)}(A)=f(A)$. Here, $^{-1}$ stands for inverse while $^{(-1)}$ stands for preimage.
My trial
Since $f:X\to Y$ is bijective, $f^{-1}:Y\to X$ exists. So, $$(f^{-1})^{(-1)}(A)=\{y\in Y:f^{-1}(y)\in A \}.$$ That's as far as I can go. Any hint or help, please?
Let $y \in Y$.
If $f^{-1}(y) \in A$, then there exists an $a \in A$ such that $f^{-1}(y) = a$. So $y = f(f^{-1}(y)) = f(a)$, and hence, $y \in f(A)$.
Conversely, if $y \in f(A)$, then there exists an $a \in A$ such that $y = f(a)$, so $f^{-1}(y) = f^{-1}(f(a)) = a$, hence $f^{-1}(y) \in A$.
Thus
$$ \{ y \in Y : f^{-1} (y) \in A \} = \{ y \in Y : f(y) \in A \} = f(A). $$