An example of why $f(f^{-1}(B))\neq B$

Question

Let $f:X\rightarrow Y$ be a function and $B\subseteq Y$ a subset of $Y$. I know (and have proven) that $f(f^{-1}(B))\subseteq B$. I've also found an example where $f(f^{-1}(B))\neq B$ for $B= \mathbb{R}$. I want to find another example, because I find my example very silly. Any help?

Answer 1

Hint: The relation $f(f^{-1}(B)) = B$ holds for all $B \subset Y$ iff $f$ is onto.

Answer 2

Let $X=Y=\Bbb R$, $f(x)=x^2$ and $B=(-1,\infty)$.

Now, $$f(f^{-1}(B))=f(\Bbb R)=[0,\infty)\subsetneq (-1,\infty).$$

Answer 3

The simplest example would be $f:\{1\}\to\{1,2\}$ with $f(1)=1$ and $B=\{2\}$.

Answer 4

Here is a way to calculate all examples, in some sense.$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} $

First, we try to simplify $\;f[f^{-1}[B]]\;$, by calculating which $\;y\;$ it contains. Implicitly, we let $\;x\;$ range over $\;X\;$.

$$\calc y \in f[f^{-1}[B]] \op\equiv\hint{definition of $\;\cdot[\cdot]\;$} \langle \exists x :: x \in f^{-1}[B] \;\land\; f(x) = y \rangle \op\equiv\hint{definition of $\;\cdot^{-1}[\cdot]\;$} \langle \exists x :: f(x) \in B \;\land\; f(x) = y \rangle \op\equiv\hint{logic: substitute RHS of $\;\land\;$ in LHS} \langle \exists x :: y \in B \;\land\; f(x) = y \rangle \op\equiv\hint{logic: extract non-$\;x\;$ clause out of $\;\exists x \;$} y \in B \;\land\;\langle \exists x :: f(x) = y \rangle \op\equiv\hint{definition of $\;\cdot[\cdot]\;$} y \in B \;\land\; y \in f[X] \endcalc$$

Now we calculate all examples:

$$\calc f[f^{-1}[B]] \not= B \op\equiv\hint{definition of set equality} \lnot \langle \forall y :: y \in f[f^{-1}[B]] \;\equiv\; y \in B \rangle \op\equiv\hint{by the previous calculation} \lnot \langle \forall y :: y \in B \land y \in f[X] \;\equiv\; y \in B \rangle \op\equiv\hint{logic: simplify} \lnot \langle \forall y :: y \in B \then y \in f[X] \rangle \op\equiv\hint{definition of $\;\subseteq\;$} \tag{*} B \not\subseteq f[X] \endcalc$$

So every $\;f,X,B\;$ that satisfy $\ref{*}$ is an example you could use.