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For any function $f:A\rightarrow B$, define a new function $g:P(A)\rightarrow P(B)$ as follows: for every $S \subseteq A$, define $g(S)=\{f(x) \mid x\in S\}$. Prove that $f$ is onto if and only if $g$ is onto.

I'm not sure how to begin, and I'm particularly confused about what exactly $g(S)$ means. Is there any insight that may help me on my way to solve this? How can I show $f$ is onto $ \leftrightarrow g$ is onto?

Thank you!

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    Draw potato pictures and arrows, and use the definition of *onto* and *powerset*..2012-10-10
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    @Berci, *potato* pictures?2012-10-10
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    Bob, $g(S)$ means exactly what it says it means. If, for example, $S=\{{3,17,{\rm dog}\}}$, then $g(S)=\{{f(3),f(17),f({\rm dog})\}}$.2012-10-10
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    How can I show that $g$ is onto if it's domain is just the power set of $A$?2012-10-10
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    Sets can be elements as well, why does it disturb you? Just suppose an element of $P(B)$ is given and find for that a preimage in $P(A)$.2012-10-10
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    How can I find the preimage in $P(A)$? By applying the definition of $g(S)$ for $P(A)$?2012-10-10
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    @GerryMyerson Ellipsoids projected onto planes, ovals, ellipses, Venn diagrams!2012-10-10

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