(c) Given sets \( A \) and \( B \) and a function from \( A \) to \( B \) as follows \[ A=\{1,2,3\}, B=\{a, b, c, d\}, f=\{(1, a),(2, a),(3, c),(2, b),(3, d)\} \] Draw an arrow diagram and determine whether the function, \( f \) is one to one or onto

To visualize the function \( f \), imagine sets \( A \) and \( B \) as two circles. In \( A \), you have the elements \( 1, 2, 3 \) and in \( B \), the elements \( a, b, c, d \). The arrows connect elements of \( A \) to their corresponding values in \( B \): - \( 1 \) points to \( a \) - \( 2 \) points to both \( a \) and \( b \) - \( 3 \) points to \( c \) and \( d \) From this representation, we can observe that \( f \) is not one-to-one (injective), since \( 2 \) maps to two different outputs (\( a \) and \( b \)), and it also shares the mapping to \( a \) with \( 1 \). However, \( f \) is not onto (surjective) either because not all elements in \( B \) are covered (specifically, \( d \) is only reached by \( 3 \), and there is no mapping for it from \( A \) alone). Hence, \( f \) is neither one-to-one nor onto.