Image and Preimage of a Set¶

Definition¶

Let $X$ and $Y$ be sets (in this course usually subsets of metric spaces), and let $f\colon X\to Y$ be a function. Let $C$ be a subset of the domain $X$ , the image of $C$ under $f$ is defined as:

f(C)=\{y\in Y\colon \text{there exists some } x\in C \text{ such that } y=f(x)\}.

Let $V$ be a subset of the codomain $Y$ , the preimage of $V$ under $f$ includes all points in the domain that give results in $V$ :

f^{-1}(V)=\{x\in X\colon f(x)\in V\}.

Interpretation¶

The image pushes a set of inputs forward: $f(C)$ collects the outputs obtained from inputs in $C$ . The preimage pulls a set of outputs back: $f^{-1}(V)$ collects all inputs whose output lands in $V$ .

Two boxes X and Y with arrows f. A set C in X is mapped onto its image f(C) in Y, and a set V in Y pulls back to its preimage, the set of points of X mapped into V.

The three marked points of $C$ are sent to only two points of $f(C)$ , so this $f$ is not injective. Going back, $f^{-1}(V)$ contains every point of $X$ that is mapped into $V$ , wherever it lies in $X$ .

A common slip is to read $f^{-1}(V)$ as requiring an inverse function. The preimage is defined for every function. For $f(x)=x^2$ on $\mathbb{R}$ , $f^{-1}(\{4\})=\{-2,2\}$ and $f^{-1}(\{-1\})=\varnothing$ $\emptyset$ . Only for a bijection does $f^{-1}(V)$ coincide with the image of $V$ under the inverse function. The same $f$ shows that the inclusions in the Key Results can be strict: for $C=[0,1]$ we get $f^{-1}(f(C))=[-1,1]$ $[- 1, 1]$ , and for $V=[-1,1]$ we get $f(f^{-1}(V))=[0,1]$ $[0, 1]$ .

Preimages behave well with set operations: they preserve unions, intersections and complements. Images preserve unions, but in general only $f(C_1\cap C_2)\subseteq f(C_1)\cap f(C_2)$ holds.

The range of $f$ is the image of the whole domain, and a level set is the preimage of a single value. In probability, the event generated by a random variable and a set $B\subseteq\mathbb{R}$ is the preimage of $B$ under the random variable.

This idea returns later: continuity can be characterized by preimages of open sets being open, whereas images of open sets need not be open.

Key Results¶

$V$ only needs to be a part of the codomain of $f$ . This means there might be some $y$ in $V$ that doesn't match any $x$ in the domain $X$ .
We have $C \subseteq f^{-1}(f(C))$ and $f(f^{-1}(V)) \subseteq V$ $V$ .

Examples¶

Let $f\colon\mathbb{R}^2\to\mathbb{R}$ be $f(x,y)=x+y$ $y$ . Image of the unit square $[0,1]\times[0,1]$ under $f$ is $[0,2]$ . Preimage of $\{0\}$ under $f$ is the line $y=-x$ $- x$ .
Let $f\colon \mathbb{R}^2\to\mathbb{R}$ be $f(x,y)=x$ $x$ . Image of the disc $\{(x,y)\in\mathbb{R}^2\colon x^2+y^2 \leq 1\}$ under $f$ is $[-1,1]$ . Preimage of $[2,3]$ under $f$ is the vertical strip $[2,3]\times\mathbb{R}$ $R$ .

Prerequisites¶

Sources¶

Math III: Analysis · Lecture 7

Keywords: image of a set, preimage, inverse image Symbols: element of, intersection, real numbers, subset

Last updated: 22 October 2026 19:04