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Images and Preimages

Discrete Mathematics Functions Basics Images and Preimages

Image of an Element

Let $f:A\rightarrow B$ be a function. The image of an element $a\in A$ is the output produced by $f$ on input $a$ ; namely $f(a)$ . If $b\in B$ is the image of an element $a\in A$ , we say that $a$ maps to $b$ , and write $a\mapsto b$ .

In other words, the image of an element $a\in A$ is nothing more than $f(a)$ ; i.e. the output of $f$ associated with $a$ .

Image of a Set

Let $f:A\rightarrow B$ be a function and let $S\subset A$ . The set $f(S)=\{f(s)\vert s\in S\}$ , i.e. the set of all outputs of $f$ on elements from $S$ , is called the image of $S$ (under $f$ ).

You can think of the image of a set $S$ to be the range of $f$ if $f$ were defined on $S$ instead of $A$ .

Quick Example Illustrating Image of an Element and Image of a Set

Let $f:A\rightarrow B$ be defined by the potato diagram below, where $A$ and $B$ are the set of elements in the left and right potatoes, respectively.

Find the image of 2 and 5 via f.
- Answer: $f(2)=4$ and $f(5)=-1$ , respectively.
Find the image of S=\{3,5,1\} via f.
- Answer: $f(S)=\{0,-1,3\}$ , i.e. the set of all stuff that $3$ , $5$ and $1$ all map to.

Preimage of an Element

Let $f:A\rightarrow B$ be a function and let $b\in B$ . The preimage of $b$ , denoted $f^{-1}[b]$ , is the set of all elements of $A$ that map to $b$ via $f$ .
Written as a set, $f^{-1}[b]=\{a\vert\ a\in A,\ \text{and}\ f(a)=b\}$ .

In other words, the preimage of an element $b$ is the set of all elements of $A$ that have $b$ as an output. Note that for preimages we use square brackets “[,]” around the argument $b$ as opposed to round braces. This is to unambiguously distinguish preimages from function inverses. Note that some authors and books may use “(,)” braces instead.

Preimage of a Set

Let $f:A\rightarrow B$ be a function and let $S\subseteq B$ . The preimage of $S$ , denoted $f^{-1}[S]$ is the set of all elements of $A$ that map to an element of $S$ via $f$ .
Written as a set: $f^{-1}[S]=\{a\vert\ a\in A,\ \text{and}\ f(a)\in S\}$

In other words, the preimage of a set $S$ is the union of all the preimages of the individual elements of $S$ .

Quick Example of Preimages of Elements and Sets

Let $g:A\rightarrow B$ be defined via the potato diagram below, where $A$ and $B$ are the set of elements in the left and right potatoes, respectively.

Find the preimages of 5 and 7
- Answer: $f^{-1}[5]=\{1,3\}$ and $f^{-1}[7]=\{6\}$ .
- Preimages of elements in the codomain set are the set of all elements in the domain set that map to the given element. In the case of $5$ , the set of elements of $A$ that map to $5\in B$ are only $1$ and $3$ . The only element in $A$ that maps to $7$ is $6$ .
Find the preimage of the set S=\{1,3,7\}
- Answer: $f^{-1}[S]=\{4, 2,5, 6\}$
- The preimage of a set $S\subseteq B$ is the set of all elements in $A$ that map to any of the elements of $S$ . So, looking at the elements of $S$ one at a time, we see that $4\in A$ is the only element that maps to $1\in S$ , that $2,5\in A$ map to $3\in S$ , and lastly that $6\in A$ maps to $7\in S$ .

Image of $5$ : $f(5)=-1$

Image of $S$ : $f(S)=\{3,4\}$

The image of a specific element in the domain set is nothing more than the output that $f$ produces from it.

The image of a set is the set of all images/outputs from each of the individual elements of the set. Since $1\mapsto 3$ and $2\mapsto 4$ , the image of the set $\{1,2\}$ is therefore $\{3,4\}$ .

$x$	$f(x)$
2.152	73
0	-1
52	0
6.8	100
1	-13
2.125	0

Image of $2.152$ : $f(2.152)=73$

Image of $K=\{52, 0,1,6.8\}$ : $f(K)=\{0,-1,-13,100\}$ .

The image of a specific element in the domain set is nothing more than the output that $f$ produces from it.

The image of a set is the set of all images/outputs from each of the individual elements of the set. Since $52\mapsto 0$ , $0\mapsto -1$ , $1\mapsto -13$ , and $6.8\mapsto 100$ the image of the set $\{52, 0,1,6.8\}$ is therefore $\{0,-1,-13,100\}$ .

Preimage of $2\in B$ : $f^{-1}[2]=\{7\}$

Preimage of $M=\{1,2,5\}$ : $f^{-1}[M]=\{4, 7, 1,3\}$

The preimage of an element $b\in B$ is the set of all elements in $A$ that map to $b$ . In our case, the only element that maps to $2\in B$ is $7$ .

The preimage of a subset of the codomain is the union of the preimages of all the elements in the given set. Put another way, the preimage of a set $S$ is the set of all elements that map to each of the elements in $S$ . In our case, we have that $4\mapsto 1$ , $7\mapsto 2$ and $1,3 \mapsto 5$ , so the preimage of the set $\{1,2,5\}$ is the set of all elements that map to $1$ , $2$ , and $5$ ; i.e. the set $\{4, 7, 1,3\}$ .

Preimage of $2$ : $\emptyset$

Preimage of $T=\{0,1\}$ : $\{1,4,8,3,5\}$

For the preimage of $2$ , there is no element that maps to a $2$ , so the preimage is empty.

For the preimage of $T=\{0,1\}$ , notice that every element in the domain maps to either $0$ or $1$ (i.e. all $x$ -values in the list of ordered pairs has a $y$ -value either $0$ or $1$ ) hence the preimage of $\{0,1\}$ (i.e. the whole codomain) is precisely the whole domain!

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