Let $f:A\to B$ be a mapping. Show that for all subsets $M$ of $A$ satisfies \begin{equation} M\subseteq f^{-1}(f(M)). \end{equation} Give an example that $f^{-1}(f(M))$ doesn't need to be equal to $M$.
Any hint where to start?
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UnknownW
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Can you add some detail about what you've done so far? – egreg Oct 29 '13 at 09:24
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Take $x \in M$, then $f(x) \in f(M)$ ... – Prahlad Vaidyanathan Oct 29 '13 at 09:24
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The equallity stands if $f$ is 1-1.If not, it may not stand. For example when $f(x)=x^2$ and $A=[-1,1]$ – Haha Oct 29 '13 at 10:57
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@egreg. Some hours later I still have no idea what to do the right way. I could only say that $x\in f^{-1}(f(M))\Leftrightarrow f(x)\in f(M)$. – UnknownW Oct 29 '13 at 19:35
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Start from the basics; when in doubt, create a new name for a complex object.
If $N\subseteq B$, then an element $x\in A$ is in $f^{-1}(N)$ if and only if $f(x)\in N$.
Set $N=f(M)$; if $x\in M$, then $f(x)\in f(M)=N$, by definition. Therefore $x\in f^{-1}(N)$. Now $f^{-1}(N)=f^{-1}(f(M))$ just by substituting symbols.
Now consider $A=\{1,2,3\}$, $B=\{4,5\}$ and define $f(1)=4$, $f(2)=4$, $f(3)=5$. Consider $M=\{1\}$. What can you say about $f^{-1}(f(M))$?
egreg
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Thanks for answering! If $f(M)=4$ since $M=\lbrace 1 \rbrace$ then $f^{-1}(f(M)) = f^{-1}(4)=1$ or $f^{-1}(4)=2$. I understand your point now. Thanks once again! – UnknownW Oct 29 '13 at 21:44
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1@AjmalW No: $f(M)={4}$; but $f^{-1}(f(M))={1,2}$, because also $f(2)\in f(M)$. – egreg Oct 29 '13 at 21:48