# 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.

#### Solutions Collecting From Web of "What is $f(f^{-1}(A))$?"

$f[f^{-1}[A]] = \{ f(x): x \in f^{-1}[A] \} = \{f(x): x \in E \text{ such that } f(x) \in A\}$, 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 $f[E] \cap A$.


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, $\;f[f^{-1}[A]] \;=\; A \cap f[E]\;$.