Bijection and image Let $f: A → B$ be a bijection, so $f^{-1}: B → A$ is a function. Let $X$ be a subset of $A$. How do I prove that $\text{Im}(f)(X)  = \text{Preim}(f^{-1})(X)$?
Thank you.
 A: See Bijection :

A function $f : A → B$ is bijective if and only if it is invertible, that is, there is a function $g: B → A$ such that $(g \circ f)$ = identity function on $A$ and $(f \circ g)$ = identity function on $B$. This function maps each image to its unique preimage.

In your case, consider for example the function $f(x) = x^2$; let $A = \mathbb Z$ (the set of  integers) and let $X = \mathbb N$ (the set of natural number). 
Clearly : $X \subseteq A$, but $f$ is not bijective.
In this case the inverse image of $\{ 4 \}$ would be $\{ -2, 2 \}$ and we have that $4 \in \mathbb N$ but $-2 \notin \mathbb N$, i.e. $-2 \notin Preim(f^{-1})(X)$.
A: Because $f$ is a bijection, we know an inverse $f^{-1}$ exists.
Let $ y \in \text{Im}_f(X) $ be arbitrary.
    \begin{align*}
        \exists x \in X. \; f(x) = y. \;\;\;\;\; &&\text{(Definition of Image)} \\
        f^{-1}(y) = x \;\;\;\;\; &&\text{(Definition of Inverse)} \\
        f^{-1}(y) \in X \;\;\;\;\; &&(x \in X)  \\
        y \in \text{PreIm}_{f^{-1}}(X) \;\;\;\;\; &&\text{(Definition of Pre-Image)}
    \end{align*}
    Since $ y \in \text{Im}_f(X) $ was arbitrary, $ \text{Im}_f(X) \subseteq \text{PreIm}_{f^{-1}}(X). \\ \\ $
Let $ y \in \text{PreIm}_{f^{-1}}(X) $ be arbitrary.
    \begin{align*}
        \exists x \in X. \; f^{-1}(y) = x. \;\;\;\;\; &&\text{(Definition of Pre-Image)} \\
        f(x) = y \;\;\;\;\; &&\text{(Definition of Inverse)} \\
        y \in \text{Im}_f(X) \;\;\;\;\; &&\text{(Definition of Image)}
    \end{align*}
    Since $ y \in \text{PreIm}_{f^{-1}}(X) $ was arbitrary, $ \text{PreIm}_{f^{-1}}(X) \subseteq \text{Im}_f(X). \\ $
By double containment, $$ \text{Im}_f(X) = \text{PreIm}_{f^{-1}}(X)  \;\;\;\;\; \blacksquare $$
