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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It's $A$ only if $f(E)\supset A$. For a set $A\subset F$, $$f^1(A):={x\in E\mid f(x)\in A}.$$ Notice that $f^{-1}(A)\subset E$, then $f\left(f^{-1}(A)\right)$ is the image of $f^{-1}(A)$. – idm Aug 31 '14 at 14:28
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@idm: I think all we need is $A \subset f(E)$. – TonyK Aug 31 '14 at 14:31
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@TonyK: I correct, thanks. – idm Aug 31 '14 at 14:33
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1The notation (which is standard) is potentially confusing. Here $f^{-1}$ does not mean the inverse function. The inverse function may not exist. We have by definition that $f^{-1}(A)$ is the set of all $x$ such that $f(x)\in A$. – André Nicolas Aug 31 '14 at 17:15
2 Answers
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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 $f[E] \cap A$.
Henno Brandsma
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$ \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, $\;f[f^{-1}[A]] \;=\; A \cap f[E]\;$.
