Prove or disprove that subsets a and b of x and f:x->y f(f-1(a))=a

[fleuryf]

Homework Statement

Prove or disprove that for any non empty sets X and Y, any subset A of X, and f:X->Y, f(f^-1(A))=A

Homework Equations

The Attempt at a Solution

I know that for x element of A f(x) = f(A) so x is an element of f^-1(f(A)). If I can assume f is injective then I can go from here but that's not given in the problem...

Note that A is a subset of X not Y.

 

[LCKurtz]

Homework Statement

Prove or disprove that for any non empty sets X and Y, any subset of X, and f:X->Y, f(f^-1(A))=A

Homework Equations

The Attempt at a Solution

I know that for x element of A f(x) = f(A) so x is an element of f^-1(f(A)). If I can assume f is injective then I can go from here but that's not given in the problem...

Your first step is to state the problem clearly and correctly. Is A a subset of X or Y? For f^-1(A) to make sense, A must be a subset of Y. But in your solution attempt you have f(A) which would only make sense of A is a subset of X.

 

[fleuryf]

Yes, turns out there was a typo in the problem as it was given which explains the difficulty in solving it.

A is indeed a subset of Y which makes the statement f(f^-1(A)) true if the mapping is surjective but in the example

X={1,3,5} Y={2,4,6,8} f:X->Y = {(1,2),(3,4),(5,6)}

Where f is not onto and A = {8}, f^-1(A) = {empty}, f(f^-1(A)) = {empty} <> A

I think this is it.

Thank you