Equivalence of Functions and Power Sets in Set Theory

[MathematicalPhysicist]

i need to prove that next three arguments are equivalent:
1)f:X->Y is on Y.
2) f:p(X)->p(Y) is on p(Y).
3)f^-1:p(Y)->p(X) is one-to-one correspondence.
where p is the power set.

 

[matt grime]

what does 'on' mean? onto? perhaps playing devils advocate a little, but mainly to make you think about the question, if f is defined on X, how is it then defined on its power set (usually denoted P(X), not p(X)).

 

[MathematicalPhysicist]

yes, i checked in mathworld, it's onto.
my main problem is with the third statement, i tried imply 3 from 1 and vice versa, but i don't know how to formualte the proof.

any further hints are appreciated.

 

[neurocomp2003]

are we assuming the function is unary?
or rather f(s in p(Y)) can be distributed into each element of s since
s is a subset of elements in Y?

 

[MathematicalPhysicist]

are we assuming the function is unary?
or rather f(s in p(Y)) can be distributed into each element of s since
s is a subset of elements in Y?

no, we don't assume it's unary.
about your second question do you mean if B is a subset of P(Y) then
f:B->f(B)={f(x)|x belongs to B} then yes, otherwise no.