Question

Given a function \(f: A \rightarrow B\) and a subset \(Y \subseteq B,\) is \(f\left(f^{-1}(Y)\right)=Y\) always true? Prove or give a counterexample.

Short answer

The statement \(f\left(f^{-1}(Y)\right) = Y\) is true if \(Y\) is a subset of the range of \(f\), but not true if \(Y\) includes elements that are not in the range of \(f\). A counterexample is given by the function \(f: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x) = x^2\) and \(Y = \{-1, 1\}.

Step-by-step solution

Understand the notation

Firstly, one must understand the notation and what the exercise asks for. \(f: A \rightarrow B\) is a function that maps elements from set \(A\) to set \(B\). \(f^{-1}(Y)\) refers to the preimage of set \(Y\) under \(f\), which is the subset of \(A\) consisting of all elements that \(f\) maps to \(Y\). \(f\left(f^{-1}(Y)\right)\) refers to then applying \(f\) to this preimage. The exercise asks whether this is equal to \(Y\) always.

Analyze the statement

The statement \(f\left(f^{-1}(Y)\right) = Y\) means that, for any element \(b\) in \(Y\), there is an element \(a\) in \(f^{-1}(Y)\) such that \(f(a) = b\). In other words, it asks if any element \(b\) in \(Y\) is an image under \(f\) of at least one element in the preimage of \(Y\). This is obviously true because \(f^{-1}(Y)\) consists of exactly those elements of \(A\) that are mapped by \(f\) to some element in \(Y\). Therefore, the statement \(f\left(f^{-1}(Y)\right) = Y\) is always true, provided that \(Y\) is a subset of the range of \(f\).

Provide a counterexample

However, if \(Y\) includes elements that are not in the range of \(f\), the statement is not true. For example, consider the function \(f: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x) = x^2\), and let \(Y = \{-1, 1\}\). Then \(f^{-1}(Y) = \{-1, 1\}\) and \(f\left(f^{-1}(Y)\right) = \{1\}\) which is not equal to \(Y\). Therefore, \(f\left(f^{-1}(Y)\right) = Y\) is not always true. It is only true if \(Y\) is a subset of the range of \(f\).