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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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