1. Given: Let f: X → Y be a function. Then we have an associated function f-1: P(Y) → P(X), where f-1 (B)⊂X is the inverse image of B⊂Y. Question: Show that f^(-1) is one-to-one if and only if f is onto. [Notes: ⊂ represents subspace, I just couldn’t find a way to put the line under the symbol. f-1 indicates the inverse of f.] 2. Ok so I know that One-to-one means we have f(x1)=f(x2), thus giving us x1=x2. And onto is ∀y∈Y, ∃x : f(x)=y. 3. What I did was just say that f-1(y1)=f-1(y2)=x thus giving me y1=y2. Or I tried proving it the other way by way of Contradiction. Saying that if f-1(y1)=f-1(y2)=x, but y1≠y2. Then we would get f(x)=y1 and f(x)=y2 but y1≠y2, thus making it not a function. So that's what I was able to come up with.