Originally Posted by

**Plato** Suppose that $\displaystyle \mathbb{G} = \left\{ {(a,f(a):a \in A} \right\}$ then define $\displaystyle \phi:A\mapsto \mathbb{G} $ as $\displaystyle \phi(a)=(a,fa)$

It is clear that $\displaystyle \phi$ must be onto.

Suppose that $\displaystyle \phi (b)= \phi (c) $ then by definition that means $\displaystyle (b,f(b))=(c,f(c))$.

By ordered pairs that means $\displaystyle c=b$ so it is one-to-one.

**The second one is false.** Let $\displaystyle D=\{a,b,c\}~\&~E=\{0,1\}$.

There are no onto functions $\displaystyle E\mapsto D$

There are no one-to-one functions $\displaystyle D\mapsto E$