5. Images and preimages

Images and preimages

Given a function f\colon X\to Y, and A\subseteq X — a subset of domain, then the set of all possible values given by f on arguments from A, we call the image A under f as is denoted as f[A]. So:

    \[f[A]=\{y\in Y\colon \exists_{a\in A} f(a)=y.\}\]

Given a subset of the set of values B\subseteq Y, the sets of all arguments such that their values are in B, is called the preimage of B under f and we denote it as f^{-1}[B]. So:

    \[f^{-1}[B]=\{x\in X\colon f(x)\in B\}.\]


Let f\colon\mathbb{N}\to\mathbb{N}, f(n)=2n. Then f[\mathbb{N}] is the set of all even numbers. If A is the set of even numbers, then f[A] is the set of all natural numbers divisible by 4. If B is the set of odd numbers, then f^{-1}[B]=\varnothing. On the other hand, f^{-1}[\{4n\colon n\in\mathbb{N}\}] is the set of even numbers.

Let now g\colon \mathbb{R}\to\mathbb{R}, g(x)=x^2. Then g[(-3,2)]=[0,9). On the other hand, g^{-1}[(-\infty,0)]=\varnothing, and g^{-1}[(-2,9)]=[0,3).

Let now h\colon \mathbb{R}^2\to\mathbb{R}, h(x,y)=xy. Then g[(-3,2)\times (-1,2]]=g[(-3,0)\times (-1,0)]\cup g[(-3,0)\times [0,2]]\cup g[[0,2)\times (-1,0)]\cup g[[0,2)\times [0,2]]=(0,3)\cup (-6,0]\cup (-2,0]\cup [0,4)=(-6,4). And g^{-1}[(-\infty,0)]=\{\left<x,y\right>\in\mathbb{R}^2\colon (x<0\land y>0)\lor (x>0\land y<0)\}.

Let finally F\colon \mathbb{R}\to\mathbb{R}^2, F(x)=\left<2x,x\right>. Therefore, F[\mathbb{R}]=\right\{\left<a,b\right>\in\mathbb{R}^2\colon b=\frac{a}{2}\left\}, so it is the line y=\frac{x}{2}. On the other hand, F^{-1}[(1,\infty)\times (1,\infty)]=\left(\frac{1}{2},\infty\right)\cap (1,\infty)=(1,\infty).