From Individual Outputs to an Image
A function assigns an output to each element of its domain. The previous tutorial studied inverse functions, which reverse the assignments of a bijection. Here we keep a function \(f:A\to B\) and ask a different question: if we select some inputs from \(A\), what collection of outputs do those selected inputs produce? The answer is another set, called the image of the selected set under \(f\).
The set being mapped must be a subset of the domain. If \(S\subseteq A\), then every \(x\in S\) is an allowed input to \(f\), and each such input has an output \(f(x)\in B\). Different elements of \(S\) may produce the same output. The image records the outputs themselves, so a repeated output appears only once as an element of the image set.
Definition (Image of a Set). Let \(f:A\to B\), and let \(S\subseteq A\). The image of \(S\) under \(f\), denoted \(f[S]\), is the set $$ f[S]=\{f(x):x\in S\}. $$ Equivalently, for any object \(y\), $$ y\in f[S] \quad\Longleftrightarrow\quad \text{there exists }x\in S\text{ such that }f(x)=y. $$
The notation \(f[S]\) makes clear that the input is a set rather than a single element. For a single input \(x\), the value \(f(x)\) is an element of \(B\). For a set \(S\), its image \(f[S]\) is a set of elements of \(B\). Some authors write \(f(S)\) for the image of a set; here we use brackets to keep the two roles visually distinct.
By definition, every element of \(f[S]\) is the output of an element of \(S\). Since \(f\) maps \(A\) into \(B\) and \(S\subseteq A\), those outputs belong to \(B\). Thus \(f[S]\subseteq B\). If \(S=A\), the image \(f[A]\) is exactly the range of \(f\), so \(f[A]=\operatorname{ran}(f)\). If \(S\) is empty, there are no inputs in \(S\) to produce outputs, and consequently \(f[\varnothing]=\varnothing\).
Computing Images Carefully
For a finite set, a direct method is to evaluate the function at each member and then collect the distinct results. For an infinite set described by a condition, the image definition asks which outputs can be written as \(f(x)\) for some \(x\) satisfying that condition. In either case, the domain restriction matters: only elements of \(S\) are used as inputs, and the resulting image remains a subset of the codomain.
Worked Example: A Finite Set with a Repeated Output
Let \(A=\{a,b,c,d\}\), \(B=\{0,1,2\}\), and define \(f:A\to B\) by $$ f(a)=1,\qquad f(b)=0,\qquad f(c)=1,\qquad f(d)=2. $$ Take \(S=\{a,c,d\}\). Evaluating \(f\) on each element of \(S\) gives \(1,1,2\). The duplicate value \(1\) is included only once in a set, so $$ f[S]=\{1,2\}. $$ In particular, \(f[S]\) is not \(\{a,c,d\}\): its elements are outputs in \(B\), not inputs in \(A\). Also, although \(S\) has three elements, its image has two. A function need not preserve the number of elements in a set.
Worked Example: An Image of a Set of Integers
Define \(f:\mathbb Z\to\mathbb Z\) by \(f(n)=3n-2\), and let $$ S=\{-2,0,3\}. $$ The three function values are $$ f(-2)=3(-2)-2=-8,\qquad f(0)=3(0)-2=-2,\qquad f(3)=3(3)-2=7. $$ Therefore $$ f[S]=\{-8,-2,7\}. $$ Each displayed output lies in the codomain \(\mathbb Z\), and each comes from an input in \(S\). Conversely, the definition of image allows no other outputs, since these are all the elements of \(S\) at which \(f\) must be evaluated.
For a set described using a variable, the image can often be simplified by expressing the outputs in a recognizable form. It is important to show both directions when claiming that two descriptions give the same set: every produced output must satisfy the proposed description, and every element of the proposed description must actually be produced.
Worked Example: Squaring a Set of Integers
Let \(q:\mathbb Z\to\mathbb Z\) be defined by \(q(n)=n^2\), and set $$ S=\{-3,-1,2\}. $$ Then $$ q(-3)=9,\qquad q(-1)=1,\qquad q(2)=4, $$ so $$ q[S]=\{1,4,9\}. $$ The inputs \(-3\) and \(3\) would have the same square, but only \(-3\) belongs to this particular \(S\). More generally, if \(T=\{-3,3\}\), then \(q[T]=\{9\}\), not a two-element set. The image collects distinct outputs and does not preserve how many inputs yield each output.
Images Preserve Inclusion and Unions
Two useful rules describe how images respond when the input set is enlarged or split into pieces. If \(S\subseteq T\subseteq A\), every input available in \(S\) is also available in \(T\), so every output produced from \(S\) is produced from \(T\). Thus images preserve inclusion. If the inputs come from \(S\cup T\), an input belongs to at least one of \(S\) or \(T\), so its output belongs to at least one of the two corresponding images. This gives an exact identity for unions.
Theorem (Images Preserve Inclusion). Let \(f:A\to B\), and let \(S,T\subseteq A\). If \(S\subseteq T\), then $$ f[S]\subseteq f[T]. $$
Proof. Suppose \(S\subseteq T\). To prove \(f[S]\subseteq f[T]\), let \(y\in f[S]\) be arbitrary. By the definition of image, there is an \(x\in S\) such that \(f(x)=y\). Since \(S\subseteq T\), this same \(x\) belongs to \(T\). Therefore \(y=f(x)\) for an \(x\in T\), which means \(y\in f[T]\). Since every element of \(f[S]\) belongs to \(f[T]\), the subset relation follows. \(\square\)
The conclusion is an inclusion, not necessarily an equality. For example, if \(f:\mathbb Z\to\mathbb Z\) is \(f(n)=n^2\), \(S=\{1\}\), and \(T=\{1,2\}\), then \(S\subseteq T\), but \(f[S]=\{1\}\) and \(f[T]=\{1,4\}\).
Theorem (Image of a Union). Let \(f:A\to B\), and let \(S,T\subseteq A\). Then $$ f[S\cup T]=f[S]\cup f[T]. $$
Proof. We prove the two inclusions. First, let \(y\in f[S\cup T]\). By the definition of image, there exists \(x\in S\cup T\) such that \(f(x)=y\). Membership in a union means \(x\in S\) or \(x\in T\). In the first case \(y\in f[S]\), and in the second case \(y\in f[T]\). Therefore \(y\in f[S]\cup f[T]\), proving \(f[S\cup T]\subseteq f[S]\cup f[T]\).
For the reverse inclusion, let \(y\in f[S]\cup f[T]\). Then \(y\in f[S]\) or \(y\in f[T]\). If \(y\in f[S]\), there is an \(x\in S\) with \(f(x)=y\); since \(S\subseteq S\cup T\), this shows \(y\in f[S\cup T]\). If \(y\in f[T]\), there is an \(x\in T\) with \(f(x)=y\); since \(T\subseteq S\cup T\), this also shows \(y\in f[S\cup T]\). Hence \(f[S]\cup f[T]\subseteq f[S\cup T]\). Equality follows by the Equality by Double Inclusion Theorem. \(\square\)
The same reasoning applies to any finite collection of subsets: the image of their union is the union of their images. In particular, the image of the empty set is empty, in agreement with the union identity. The essential fact is that membership in a union is an “or” condition, and an input from either part contributes its output to the corresponding image.
Worked Example: Using the Union Identity
Let \(h:\mathbb Z\to\mathbb Z\) be \(h(n)=n+4\), with \(S=\{-4,0\}\) and \(T=\{2,5\}\). Direct evaluation gives $$ h[S]=\{0,4\},\qquad h[T]=\{6,9\}. $$ The theorem predicts $$ h[S\cup T]=h[S]\cup h[T]=\{0,4,6,9\}. $$ Checking the inputs in \(S\cup T=\{-4,0,2,5\}\) yields \(h(-4)=0\), \(h(0)=4\), \(h(2)=6\), and \(h(5)=9\), confirming the predicted image.
Intersections Require Care
For intersections, one inclusion always holds: $$ f[S\cap T]\subseteq f[S]\cap f[T]. $$ If \(y\in f[S\cap T]\), it comes from an input \(x\) that belongs to both \(S\) and \(T\). Consequently \(y\) belongs to both images. The reverse inclusion can fail. An output may belong to \(f[S]\) because of one input and to \(f[T]\) because of a different input; those two inputs need not be in \(S\cap T\).
Worked Example: The Image of an Intersection Can Be Strictly Smaller
Let \(q:\mathbb Z\to\mathbb Z\) be \(q(n)=n^2\), and choose \(S=\{-2\}\) and \(T=\{2\}\). These input sets are disjoint, so $$ S\cap T=\varnothing \quad\text{and}\quad q[S\cap T]=q[\varnothing]=\varnothing. $$ However, \(q[S]=\{4\}\) and \(q[T]=\{4\}\). Therefore $$ q[S]\cap q[T]=\{4\}. $$ Thus \(q[S\cap T]\) is a proper subset of \(q[S]\cap q[T]\). The output \(4\) has been produced from \(-2\) in \(S\) and from \(2\) in \(T\), even though neither input lies in both sets.
Injectivity prevents this particular mismatch of inputs. If \(f\) is injective and an output \(y\) lies in both \(f[S]\) and \(f[T]\), then there are \(s\in S\) and \(t\in T\) with \(f(s)=y=f(t)\). Injectivity gives \(s=t\), so this one input lies in \(S\cap T\). As a result, for an injective function the intersection formula is an equality. For a function that is not injective, only the inclusion is guaranteed.
Proposition (Images of Intersections under an Injective Function). Let \(f:A\to B\) be injective, and let \(S,T\subseteq A\). Then $$ f[S\cap T]=f[S]\cap f[T]. $$
Proof. First, if \(y\in f[S\cap T]\), then \(y=f(x)\) for some \(x\in S\cap T\). Since \(x\in S\) and \(x\in T\), we have \(y\in f[S]\) and \(y\in f[T]\). Thus \(f[S\cap T]\subseteq f[S]\cap f[T]\).
For the reverse inclusion, let \(y\in f[S]\cap f[T]\). There are \(s\in S\) and \(t\in T\) such that \(f(s)=y=f(t)\). Because \(f\) is injective, \(s=t\). This common input therefore belongs to both \(S\) and \(T\), so \(s\in S\cap T\). Since \(f(s)=y\), the definition of image gives \(y\in f[S\cap T]\). Hence \(f[S]\cap f[T]\subseteq f[S\cap T]\), and the two inclusions prove equality. \(\square\)
A Reliable Method for Image Problems
When the set or function is given by a formula, the definition provides a dependable route to a solution. Start with the inputs in the specified set, determine their outputs, and then describe exactly the set of resulting values. For an identity between image sets, prove both inclusions or use a previously proved equality. Keep track of where the variables belong: \(x\) is selected from a subset of the domain, while \(f(x)\) belongs to the codomain.
The range is one important special case of an image: it is the image of the entire domain. Images also allow a function to be studied on just part of its domain without changing the function itself. In later work, the direction of the membership question will be reversed: instead of asking which outputs come from a chosen set of inputs, we will ask which inputs map into a chosen set of outputs. Keeping these two questions distinct is essential for understanding how functions act on sets.
Check Your Understanding
- Let \(f:A\to B\) and \(S\subseteq A\). State the definition of \(f[S]\), and explain why \(f[S]\subseteq B\).
- Suppose \(f(u)=2\), \(f(v)=2\), and \(f(w)=5\), with \(S=\{u,v,w\}\). What is \(f[S]\), and why does it have fewer elements than \(S\)?
- If \(S\subseteq T\subseteq A\), what inclusion must hold between \(f[S]\) and \(f[T]\)?
- State the identity relating \(f[S\cup T]\) to \(f[S]\) and \(f[T]\).
- For a general function, which inclusion always relates \(f[S\cap T]\) and \(f[S]\cap f[T]\)? What additional hypothesis guarantees equality?