Relations and functions · Definition

Image and preimage

set theoryrelations and functions

Before you start you need to know

Image and preimage are the two ways of moving subsets through a function: one goes forward from the domain to the codomain, and the other goes back.

Definition

Given \(f\colon A\to B\), the image of \(X\subseteq A\) is the set of values \(f(x)\) with \(x\in X\), and the preimage of \(Y\subseteq B\) is the set of \(a\in A\) whose value lies in \(Y\).

\[f(X)=\{f(x)\colon x\in X\},\qquad f^{-1}(Y)=\{a\in A\colon f(a)\in Y\}\]

The image of \(X\subseteq A\) under \(f\) is \(f(X)=\{f(x)\colon x\in X\}\), a subset of \(B\). In particular \(f(A)\) is the image of the function, and \(f\) is surjective exactly when \(f(A)=B\). For \(f\colon\mathbb{R}\to\mathbb{R}\), \(f(x)=x^2\), we have \(f([-1,2])=[0,4]\).

The preimage of \(Y\subseteq B\) is \(f^{-1}(Y)=\{a\in A\colon f(a)\in Y\}\). For the same example, \(f^{-1}([1,4])=[-2,-1]\cup[1,2]\). The notation \(f^{-1}(Y)\) does not presuppose that \(f\) has an inverse: the preimage exists for every function, even when \(f^{-1}(\{y\})\) has several elements or none.

Typical mistakes: assuming \(f(f^{-1}(Y))=Y\) always. What always holds is \(f(f^{-1}(Y))\subseteq Y\), with equality only when \(Y\subseteq f(A)\). With \(f(x)=x^2\), \(f(f^{-1}([-1,4]))=[0,4]\neq[-1,4]\).

Think about it

Think about it: why can f(f^{-1}(Y)) be smaller than Y?

Because f(f^{-1}(Y)) only contains values that f actually takes and that lie in Y; points of Y outside f(A) are never reached.

Exercises 10

References

  • Herbert B. Enderton, Elements of Set Theory, 1st edition, Academic Press, 1977.
  • Thomas Jech, Set Theory (The Third Millennium Edition), 3rd millennium edition, Springer, 2003.