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.
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 \(f^{-1}[\emptyset]=\emptyset\).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:
\(f^{-1}[\{4,6\}]=\{3\}\),
\(f^{-1}[\{4,5\}]=\{1,2,3\}=X\),
\(f^{-1}[\{6\}]=\emptyset\),
\(f^{-1}[\{5\}]=\{1,2\}\).
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:
\[f^{-1}[(0,1)]=\left\{\left(\tfrac{5}{2}y+\varepsilon,y\right):\ \epsilon,y\in\mathbb{R},\ 0<\varepsilon<\tfrac{1}{2} \right\}.\]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.
Consider \(f:X\to Y\) and let \(U,V\subseteq Y.\) Then
\(f^{-1}[U\cap V]=f^{-1}[U]\cap f^{-1}[V]\);
\(f^{-1}[U\cup V]=f^{-1}[U]\cup f^{-1}[V].\)
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.
Consider \(f:X\to Y\), and let \(U\subseteq X\) and \(V\subseteq Y\). Then
\(f[f^{-1}[V]]\subseteq V\), and for \(f\) surjective, we have \(f[f^{-1}[V]]=V\).
\(U\subseteq f^{-1}[f[U]]\), and for \(f\) injective, we have \(U=f^{-1}[f[U]].\)
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.□