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.]
To show that f^(-1) is one-to-one if and only if f is onto, we need to prove both implications separately.
First, let's prove that if f^(-1) is one-to-one, then f is onto.
Assume that f^(-1) is one-to-one. This means that for any two sets B1 and B2 in P(Y) such that B1 ≠ B2, we have f^(-1)(B1) ≠ f^(-1)(B2).
To prove that f is onto, we need to show that for every element y in Y, there exists an element x in X such that f(x) = y.
Let's choose an arbitrary element y in Y. We want to find an x in X such that f(x) = y.
Consider the set B = {y}. Since f^(-1) is one-to-one, f^(-1)(B) ≠ f^(-1)(∅), where ∅ represents the empty set.
Now, f^(-1)(B) ≠ f^(-1)(∅) implies that there exists an x' in X such that f(x') ≠ ∅. This means there exists an x' in X such that f(x') = y.
Since we've chosen an arbitrary y in Y and shown that there exists an x in X such that f(x) = y, we conclude that f is onto.
Next, let's prove that if f is onto, then f^(-1) is one-to-one.
Assume that f is onto. This means that for every element y in Y, there exists an element x in X such that f(x) = y.
To prove that f^(-1) is one-to-one, we need to show that for any two sets B1 and B2 in P(Y) such that B1 ≠ B2, we have f^(-1)(B1) ≠ f^(-1)(B2).
Let's consider two arbitrary sets B1 and B2 in P(Y) such that B1 ≠ B2. We want to show that f^(-1)(B1) ≠ f^(-1)(B2).
Since B1 ≠ B2, there must exist at least one y1 in B1 that is not in B2, or vice versa.
Without loss of generality, let's assume that there exists a y1 in B1 that is not in B2.
Since f is onto, there exists an x1 in X such that f(x1) = y1. This means that x1 is in f^(-1)(B1).
Now, suppose f^(-1)(B1) = f^(-1)(B2). This means that for every x in f^(-1)(B1), we have f(x) in B2.
However, we've already shown that there exists an x1 in f^(-1)(B1) (namely, x1) such that f(x1) = y1, which is not in B2. This contradicts our assumption that f^(-1)(B1) = f^(-1)(B2).
Hence, we conclude that f^(-1)(B1) ≠ f^(-1)(B2) for any two sets B1 and B2 in P(Y) such that B1 ≠ B2.
Therefore, f^(-1) is one-to-one.
Overall, we've proven both implications. If f^(-1) is one-to-one, then f is onto. Conversely, if f is onto, then f^(-1) is one-to-one.