# Playing with images and inverse images

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