Why Images Do Not Preserve Complements

Two Complements, in Two Different Sets

For a function \(f:X\to Y\) and a subset \(A\subseteq X\), there are two natural sets to compare: the image of the part of the domain outside \(A\), and the part of the codomain outside the image of \(A\). They are \(f(X\setminus A)\) and \(Y\setminus f(A)\). The first records outputs that actually come from points outside \(A\); the second records all codomain elements that are not outputs of points in \(A\). These descriptions need not agree.

For clarity, the image of \(C\subseteq X\) is \(f(C)=\{f(x):x\in C\}\). The complement \(X\setminus A\) is taken in the domain, while \(Y\setminus f(A)\) is taken in the codomain. In particular, elements of \(Y\) that are not outputs of \(f\) can belong to \(Y\setminus f(A)\), but they cannot belong to \(f(X\setminus A)\).

What Can Be Said in General?

Proof. Let \(y\in f(X)\setminus f(A)\). Since \(y\in f(X)\), there is some \(x\in X\) with \(f(x)=y\). Since \(y\notin f(A)\), this \(x\) cannot lie in \(A\): if it did, \(y=f(x)\) would belong to \(f(A)\). Thus \(x\in X\setminus A\), so \(y=f(x)\in f(X\setminus A)\). This proves the inclusion. \(\square\)

The proposition is about \(f(X)\setminus f(A)\), not necessarily \(Y\setminus f(A)\). If \(f\) is not surjective, the latter set may contain elements outside \(f(X)\). Even when \(f\) is surjective, the inclusion can be strict: an output may come both from a point in \(A\) and from a point outside \(A\). The next examples separate these two reasons for failure.

When Does Equality Hold?

For one particular subset \(A\), equality has two separate requirements. First, no output from outside \(A\) can also be the output of a point in \(A\). Second, every codomain element outside \(f(A)\) must actually be reached by a point outside \(A\).

Proof. Suppose first that \(f(X\setminus A)=Y\setminus f(A)\). If \(x\in f^{-1}(f(A))\), then \(f(x)\in f(A)\). If \(x\notin A\), then \(f(x)\in f(X\setminus A)=Y\setminus f(A)\), contradicting \(f(x)\in f(A)\). Hence \(x\in A\), proving \(f^{-1}(f(A))\subseteq A\). Also every element of \(Y\setminus f(A)=f(X\setminus A)\) belongs to \(f(X)\), proving the second condition.

Conversely, suppose both conditions hold. If \(y\in f(X\setminus A)\), then \(y=f(x)\) for some \(x\notin A\). If \(y\) belonged to \(f(A)\), then \(x\in f^{-1}(f(A))\subseteq A\), a contradiction. Thus \(y\in Y\setminus f(A)\). This gives \(f(X\setminus A)\subseteq Y\setminus f(A)\). Now let \(y\in Y\setminus f(A)\). The second condition gives \(y\in f(X)\), so \(y=f(x)\) for some \(x\in X\). If \(x\in A\), then \(y\in f(A)\), which is impossible. Therefore \(x\in X\setminus A\), and \(y\in f(X\setminus A)\). This proves the reverse inclusion and the equality. \(\square\)

The first condition rules out collisions across the split between \(A\) and \(X\setminus A\). It does not require that \(f\) be injective everywhere: two points both in \(A\), for example, may have the same output without violating it. The second condition rules out missing outputs in the part of the codomain being compared.

Proof. Suppose \(f\) is bijective and let \(A\subseteq X\). If \(y\in f(X\setminus A)\), then \(y=f(x)\) for some \(x\notin A\). If \(y\in f(A)\), then \(y=f(a)\) for some \(a\in A\). Injectivity would imply \(x=a\), contradicting \(x\notin A\). Therefore \(f(X\setminus A)\subseteq Y\setminus f(A)\). Conversely, if \(y\in Y\setminus f(A)\), surjectivity gives \(y=f(x)\) for some \(x\in X\). This \(x\) cannot lie in \(A\), since that would give \(y\in f(A)\). Thus \(x\in X\setminus A\), so \(y\in f(X\setminus A)\). Equality follows for every \(A\).

Now suppose the equality holds for every subset \(A\subseteq X\). Taking \(A=\varnothing\) gives \(f(X)=Y\), because \(f(\varnothing)=\varnothing\) and \(X\setminus\varnothing=X\). Hence \(f\) is surjective. To prove injectivity, let \(x,x'\in X\) satisfy \(f(x)=f(x')\). If \(x\ne x'\), take \(A=\{x\}\). Then \(x'\in X\setminus A\), so \(f(x')\in f(X\setminus A)\). But \(f(x')=f(x)\in f(A)\), whereas the assumed equality gives \(f(X\setminus A)=Y\setminus f(A)\). No element can belong to both \(f(A)\) and \(Y\setminus f(A)\), a contradiction. Thus \(x=x'\), and \(f\) is injective. The argument also covers an empty domain: the equality for \(A=\varnothing\) forces \(Y=f(X)=\varnothing\), so the unique map from the empty set to the empty set is bijective. \(\square\)

A Common Mistake and a Reliable Test

Common pitfall. A tempting but incorrect calculation is to write \(f(X\setminus A)=Y\setminus f(A)\) simply by “taking images of both sides” of \(X\setminus A\). The example with \(X=\{1,2,3\}\), \(Y=\{a,b\}\), \(f(1)=a\), \(f(2)=a\), \(f(3)=b\), and \(A=\{1\}\) gives a direct correction: \(f(X\setminus A)=\{a,b\}\), while \(Y\setminus f(A)=\{b\}\). The image operation does not remove the output \(a\) from the first set just because \(a\) also occurs as the image of a point in \(A\).

When comparing these sets, compute them separately: first list the domain points outside \(A\) and take their outputs; then find \(f(A)\) and remove those outputs from all of \(Y\). Keep track of whether the issue is a shared output or an element of \(Y\) that is not in \(f(X)\). For a bijection neither issue occurs, and the theorem proves the equality for every subset.