Find a function such that where is the image of f at E and 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[ ( 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[ ( R(f))]= R(f) then f is onto.
So any function which is not onto will satisfy f[ ( R(f))]=/= R(f).
Now to prove the above .
But to prove that f is onto if f[ ( 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:
[ yεR(f) -------> ( xεD(f) & y=f(x))].
Let yεR(f),since f[ ( R(f))]= R(f),then yεf[ ( R(f))].
But yεf[ ( R(f))] <=====> y= f(x) & xε[ ( R(f))]..
But xε[ ( R(f))] <====> xεD(f) & f(x)εR(f).
Hence; yεf[ ( 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