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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.

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

$

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

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