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 two statements:
1. If f is onto, then f^(-1) is one-to-one.
2. If f^(-1) is one-to-one, then f is onto.
Let's start by proving the first statement:
1. If f is onto, then f^(-1) is one-to-one.
To prove this, we need to show that for any two subsets B1 and B2 in the power set P(Y), if f^(-1)(B1) = f^(-1)(B2), then B1 = B2.
Let's assume that f is onto. That means for every element y in Y, there exists at least one element x in X such that f(x) = y.
Now, let's consider two arbitrary subsets B1 and B2 in P(Y) such that f^(-1)(B1) = f^(-1)(B2).
This means that for any x in X, x belongs to f^(-1)(B1) if and only if x belongs to f^(-1)(B2).
By the definition of f^(-1), this implies that for any x in X, f(x) belongs to B1 if and only if f(x) belongs to B2.
Since f is onto, for any y in Y, there exists at least one x in X such that f(x) = y. Therefore, for any y in Y, y belongs to B1 if and only if y belongs to B2.
By the definition of subsets, this means that B1 = B2.
Thus, we have shown that if f is onto, then f^(-1) is one-to-one.
Now, let's move on to proving the second statement:
2. If f^(-1) is one-to-one, then f is onto.
To prove this, we need to show that for every y in Y, there exists at least one x in X such that f(x) = y.
Let's assume that f^(-1) is one-to-one. That means for any two subsets B1 and B2 in P(Y), if f^(-1)(B1) = f^(-1)(B2), then B1 = B2.
Now, let's consider an arbitrary y in Y. We want to show that there exists at least one x in X such that f(x) = y.
Since f^(-1) is one-to-one, let's consider two subsets B1 and B2 in P(Y) defined as follows:
B1 = {y}
B2 = Y - {y}
By the definition of inverse image, we have:
f^(-1)(B1) = {x ∈ X | f(x) ∈ B1} = {x ∈ X | f(x) = y}
f^(-1)(B2) = {x ∈ X | f(x) ∈ B2} = {x ∈ X | f(x) ≠ y}
Since B1 and B2 are distinct subsets in P(Y), f^(-1)(B1) and f^(-1)(B2) must also be distinct subsets in P(X) due to the one-to-one property of f^(-1).
Therefore, there exists at least one x in X such that f(x) = y.
Since y was arbitrary, this holds for every y in Y.
Hence, we have shown that if f^(-1) is one-to-one, then f is onto.
By proving both statements, we have established that f^(-1) is one-to-one if and only if f is onto.