Find a function $\displaystyle f: D \to E$ such that $\displaystyle f[f^{-1}[E]] \neq E$ where $\displaystyle f[E]$ is the image of f at E and $\displaystyle f^{-1}[E]$ is the pre-image of f at E.
Suppose that $\displaystyle D = \left\{ {1,2,3} \right\}\,\& \,E = \left\{ {a,b,c} \right\}$.
Define a function
$\displaystyle \begin{gathered} f = \left\{ {\left( {1,a} \right),\left( {2,c} \right),\left( {3,a} \right)} \right\} \hfill \\
f^{ - 1} (E) = D \hfill \\ f\left( {f^{ - 1} (E)} \right) = f\left( D \right) = \left\{ {a,c} \right\} \ne E \hfill \\ \end{gathered} $
The theorem covering the above is:
For all functions f : ,if f is not onto then f[$\displaystyle f^-1$ ( 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[$\displaystyle f^-1$ ( R(f))]= R(f) then f is onto.
So any function which is not onto will satisfy f[$\displaystyle f^-1$ ( R(f))]=/= R(f).
Now to prove the above .
But to prove that f is onto if f[$\displaystyle f^-1$ ( 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:
$\displaystyle \forall y$[ yεR(f) ------->$\displaystyle \exists x$( xεD(f) & y=f(x))].
Let yεR(f),since f[$\displaystyle f^-1$ ( R(f))]= R(f),then yεf[$\displaystyle f^-1$ ( R(f))].
But yεf[$\displaystyle f^-1$ ( R(f))] <=====> y= f(x) & xε[$\displaystyle f^-1$ ( R(f))]..
But xε[$\displaystyle f^-1$ ( R(f))] <====> xεD(f) & f(x)εR(f).
Hence; yεf[$\displaystyle f^-1$ ( 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