We’re looking here to relational expressions involving image and inverse image.

We consider a function \(f : X \to Y\) from the set \(X\) to the set \(Y\). If \(x\) is a member of \(X\), \(f(x)\) is the image of \(x\) under \(f\).

The **image** of a subset \(A \subset X\) under \(f\) is the subset (of \(Y\)) \[f(A)\stackrel{def}{=} \{f(x) : x \in A\}.\]

The **inverse image** of a subset \(B \subset Y\) is the subset of \(A\) \[f^{-1}(B)\stackrel{def}{=} \{x \in X : f(x) \in B\}.\] *Important to understand is that here, \(f^{-1}\) is not the inverse function of \(f\).
*

We now look at relational expressions involving the

**image**and the

**inverse image**under \(f\).

### Inverse images with unions and intersections

Following relations hold:

\[\begin{array}{c}

f^{-1}(B_1 \cup B_2) = f^{-1}(B_1) \cup f^{-1}(B_2)\\

f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)

\end{array}\] Let’s prove the first equality as an example. We have \(x \in f^{-1}(B_1 \cup B_2)\) if and only if \(f(x) \in B_1 \cup B_2\) if and only if \(f(x) \in B_1\) or \(f(x) \in B_2\) which means exactly \(x \in f^{-1}(B_1) \cup f^{-1}(B_2)\). Continue reading Playing with images and inverse images