1. Given: Let f: X → Y be a function. Then we have an associated function f(adsbygoogle = window.adsbygoogle || []).push({}); ^{-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(x_{1})=f(x_{2}), thus giving us x_{1}=x_{2}. And onto is ∀y∈Y, ∃x : f(x)=y.

3. What I did was just say that f^{-1}(y_{1})=f^{-1}(y_{2})=x thus giving me y_{1}=y_{2}. Or I tried proving it the other way by way of Contradiction. Saying that if f^{-1}(y_{1})=f^{-1}(y_{2})=x, but y_{1}≠y_{2}. Then we would get f(x)=y_{1}and f(x)=y_{2}but y_{1}≠y_{2}, thus making it not a function. So that's what I was able to come up with.

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# Introduction to Proofs: One-to-One and Onto Problem

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