Q.Given $A=\{1,2,3\}$ and $B=\{4,5,6\}$, and the relation $R=\{(1,4),(2,5),(3,6)\}$ from $A$ to $B$, show that $R$ is a function and state its domain, codomain and range.

A relation from $A$ to $B$ is a function only if every element of $A$ has exactly one image in $B$, which is exactly what needs to be checked here.

Every element of $A$ appears exactly once as a first component, so no element is left unpaired and no element is paired twice — the defining condition for a function.

$R=\{(1,4),(2,5),(3,6)\}$ is a relation from $A=\{1,2,3\}$ to $B=\{4,5,6\}$, since $R \subseteq A \times B$.

To check whether $R$ is a function, we verify that every element of $A$ is paired with exactly one element of $B$:

• $1$ is paired with $4$ only.
• $2$ is paired with $5$ only.
• $3$ is paired with $6$ only.

Every element of $A$ appears exactly once as a first component of an ordered pair in $R$ — no element of $A$ is missing, and no element of $A$ is repeated with two different images. This satisfies the definition of a function, so $R$ is indeed a function from $A$ to $B$, and we may write $f: A \to B$ with $f(1)=4,\ f(2)=5,\ f(3)=6$.

• Domain $= A = \{1,2,3\}$, since every element of $A$ has an image.
• Codomain $= B = \{4,5,6\}$, the set the images are drawn from.
• Range $= \{f(1),f(2),f(3)\} = \{4,5,6\}$ — here the range happens to equal the codomain, so this particular function is also onto.

Students often confuse a general relation with a function and skip checking that every element of $A$ (not just some of them) has an image. Another common slip is stating the range as equal to the codomain by assumption rather than by actually listing the images obtained.