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    Super Member Aryth's Avatar
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    Functions

    Find a function f: D \to E such that f[f^{-1}[E]] \neq E where f[E] is the image of f at E and f^{-1}[E] is the pre-image of f at E.
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    Suppose that D = \left\{ {1,2,3} \right\}\,\& \,E = \left\{ {a,b,c} \right\}.
    Define a function
    \begin{gathered}  f = \left\{ {\left( {1,a} \right),\left( {2,c} \right),\left( {3,a} \right)} \right\} \hfill \\<br />
  f^{ - 1} (E) = D \hfill \\  f\left( {f^{ - 1} (E)} \right) = f\left( D \right) = \left\{ {a,c} \right\} \ne E \hfill \\ \end{gathered}
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    Quote Originally Posted by Aryth View Post
    Find a function f: D \to E such that f[f^{-1}[E]] \neq E where f[E] is the image of f at E and f^{-1}[E] is the pre-image of f at E.
    The theorem covering the above is:


    For all functions f : ,if f is not onto then f[  f^-1 ( R(f))]=/= R(f) ..........where R(f) is the range of f.

    But due to contrapositive law the above theorem is equivalent to:

    For all functions f : if f[  f^-1 ( R(f))]= R(f) then f is onto.

    So any function which is not onto will satisfy f[  f^-1 ( R(f))]=/= R(f).


    Now to prove the above .

    But to prove that f is onto if f[  f^-1 ( R(f))]= R(f),we must prove that:


    For all yεR(f) then there exists an xεD(f) such that y=f(x),where D(f)is the domain of f .

    The above in symbols is:

    \forall y[ yεR(f) -------> \exists x( xεD(f) & y=f(x))].


    Let yεR(f),since f[  f^-1 ( R(f))]= R(f),then yεf[  f^-1 ( R(f))].

    But yεf[  f^-1 ( R(f))] <=====> y= f(x) & xε[  f^-1 ( R(f))]..




    But xε[  f^-1 ( R(f))] <====> xεD(f) & f(x)εR(f).


    Hence; yεf[  f^-1 ( R(f))]<=======> y=f(x) & xεD(f) & f(x)εR(f).

    Thus we see that there an xεD(f) such that y= f(x), so f is onto
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