6.3 Pre-images

As well as looking at the image of various subset of X under the function f:X\to Y, we are often interested in looking at where certain subsets of Y come from.

Definition 6.13:

Consider f:X\to Y and let V\subseteq Y. We define the pre-image (or inverse image) of V under f by f^{-1}[V]=\{x\in X:\ f(x)\in V \}.

We have f^{-1}[\emptyset]=\emptyset.
Example:

Let X=\{1,2,3\} and Y=\{4,5,6\}. Define the function f=\{(1,5),(2,5),(3,4)\} \subseteq X\times Y. Then:

  • f^{-1}[\{4,6\}]=\{3\},

  • f^{-1}[\{4,5\}]=\{1,2,3\}=X,

  • f^{-1}[\{6\}]=\emptyset,

  • f^{-1}[\{5\}]=\{1,2\}.

Example:

Let f:\mathbb{R}\times \mathbb{R}\to\mathbb{R} be defined by f((x,y))=2x-5y. We will find f^{-1}[\{0\}] and f^{-1}[(0,1)] (recall that the open interval (0,1)=\{x\in\mathbb{R}:0<x<1\}).

\begin{align*} f^{-1}[\{0\}]&=\{(x,y)\in\mathbb{R}\times\mathbb{R}:\ f((x,y))\in\{0\} \}\\ &=\{(x,y)\in\mathbb{R}\times\mathbb{R}:\ 2x-5y=0 \}\\ &=\{(x,y)\in\mathbb{R}\times\mathbb{R}: \ y=\tfrac{2x}{5} \}\\ &=\left\{\left(x,\tfrac{2x}{5}\right):\ x\in\mathbb{R} \right\}. \end{align*}

\begin{align*} f^{-1}[(0,1)]&=\{(x,y)\in\mathbb{R}\times\mathbb{R}:\ f((x,y))\in (0,1) \}\\ &=\{(x,y)\in\mathbb{R}\times\mathbb{R}:\ 2x-5y\in(0,1)\}\\ &=\{(x,y)\in\mathbb{R}\times\mathbb{R}:\ 0<2x-5y<1 \}\\ &=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}:\ \tfrac{5}{2}y<x<\tfrac{5}{2}y+\tfrac{1}{2} \right\}. \end{align*}

We can also rearrange the last line to say:

f^{-1}[(0,1)]=\left\{\left(\tfrac{5}{2}y+\varepsilon,y\right):\ \epsilon,y\in\mathbb{R},\ 0<\varepsilon<\tfrac{1}{2} \right\}.
Example:

Define g:\mathbb{R}\to \mathbb{R} by g(x)=x^2. Take V=[4,\infty)=\{x\in \mathbb{R}: 4\leq x\}. Then

\begin{align*} g^{-1}[V]&=\{x\in\mathbb{R}:\ g(x)\in V \}\\ &=\{x\in\mathbb{R}:\ x^2\in [4,\infty) \}\\ &=\{x\in\mathbb{R}:\ (x\ge 2)\ \vee\ (x\le -2) \}\\ &=(-\infty,-2 ]\cup[2,\infty). \end{align*}

We have the nice results that union and intersection behave as expected.

Theorem 6.14:

Consider f:X\to Y and let U,V\subseteq Y. Then

  1. f^{-1}[U\cap V]=f^{-1}[U]\cap f^{-1}[V];

  2. f^{-1}[U\cup V]=f^{-1}[U]\cup f^{-1}[V].

Proof.

We prove a. and leave b. as an exercise.

We have \begin{align*} f^{-1}[U\cap V]&=\{x\in X:\ f(x)\in U\cap V\ \}\\ &=\{x\in X:\ f(x)\in U \wedge f(x)\in V\ \}\\ &=\{x\in X:\ f(x)\in U\ \}\cap\{x\in X:\ f(x)\in V\ \}\\ &=f^{-1}[U]\cap f^{-1}[V]. \end{align*}

Therefore the two sets are equal, i.e. f^{-1}[U\cap V]=f^{-1}[U]\cap f^{-1}[V].

The link between images and pre-images is explored in the following theorem.

Theorem 6.15:

Consider f:X\to Y, and let U\subseteq X and V\subseteq Y. Then

  1. f[f^{-1}[V]]\subseteq V, and for f surjective, we have f[f^{-1}[V]]=V.

  2. U\subseteq f^{-1}[f[U]], and for f injective, we have U=f^{-1}[f[U]].

Proof.

We prove a. and leave b. as an exercise.

If V=\emptyset, then f^{-1}[V]=\emptyset and f[f^{-1}[V]]=\emptyset=V. So we assume V\neq\emptyset. Similarly, as \emptyset\subseteq V, we have that if f[f^{-1}[V]]=\emptyset then f[f^{-1}[V]]\subseteq V. So we also assume f[f^{-1}[V]]\neq\emptyset.

Suppose that y\in f[f^{-1}[V]]. Then y=f(w) for some w\in f^{-1}[V]. By the definition of f^{-1}[V], we have f(w)\in V. Hence, y=f(w)\in V. Since y is arbitrary, this shows that every element of f[f^{-1}[V]] lies in V, that is f[f^{-1}[V]]\subseteq V.

Now suppose that f is surjective. We need to show that V\subseteq f[f^{-1}[V]]. Suppose that v\in V. Since f is surjective, there exists an x\in X such that f(x)=v. Then f(x)\in V, and it follows that x\in f^{-1}[V]. Hence, v=f(x)\in f[f^{-1}[V]]. Since v was arbitrary, this shows that V\subseteq f[f^{-1}[V]]. Summarising, we have that f[f^{-1}[V]]=V whenever f is surjective.