For every function f, subset A of the domain and subset B of the codomain prove that If f is injective we have A = f −1(fA) and if f is surjective we have f(f −1B) = B.
It should read for the first one
(this is always true). So,
Proof: Suppose that and let . Since there exists some such that . But, since we see that and so which condtradicts 's injectivity. The conclusion follows
from where it follows that
Let then and so from where it follows that