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### Problem 3

**Solution**

Let $A=\{1,2,3\}$ and $B=\{2,4,6\}$ . For each relation between $A$ and $B$ given as a subset of $A \times B$, decide whether it is a function mapping $A$ into $B$. If it is a function, decide whether it is one-to-one and whether it is onto $B$. Justify your answers.

(a) $\phi = \{(2,4),(3,6),(2,2)\}$

$\phi$ is NOT a function mapping $A$ into $B$ because each element of $A$ does not appear as the first member of exactly one ordered pair in $\phi$. Specifically, the element $2$ appears more than once as the first member of an ordered pair in $\phi$, while the element $1$ does not appear at all.

(b) $\phi = \{(1,4),(2,6),(3,2)\}$

$\phi$ is a function mapping $A$ into $B$ because each element of $A$ appears as the first element of exactly one ordered pair in $\phi$.

$\phi$ is one-to-one because each element in $A$ maps to exactly one element in $B$.

$\phi$ is onto $B$ because the range of $\phi$, $\{4,6,2\}$, is equal to the codomain, $B$.

(c) $\phi = \{(1,4),(3,6),(2,4)\}$

$\phi$ is a function mapping $A$ into $B$ because each element of $A$ appears as the first element of exactly one ordered pair in $\phi$.

$\phi$ is NOT onto $B$ because the codomain, $B$, is not equal to the range, $\{4,4,6\}$.

$\phi$ is NOT one-to-one because it maps two elements from $A$ ($1$ and $2$) to the same element ($4$) from $B$.