1. ## Graphs

Given $\displaystyle f: X \to Y$ is a function with graph $\displaystyle G_f \subseteq X \times Y$ prove that $\displaystyle f$ is surjective $\displaystyle \Leftrightarrow \forall y \in Y, (X \times \{y \} \cap G_f) \neq \emptyset$.

Proof: '$\displaystyle \Rightarrow$' direction: If $\displaystyle f$ is surjective, then $\displaystyle \forall y \in Y, \exists x \in X, y = f(x)$. Then $\displaystyle X \times \{y \} \cap G_f \Rightarrow x \in X$ and $\displaystyle y \in \{y \}$ or $\displaystyle y = y$. Then $\displaystyle X \times \{y \} \cap G_f \neq \emptyset$ because by definition $\displaystyle f$ is surjective.

'$\displaystyle \Leftarrow$' direction: If $\displaystyle \forall y \in Y, (X \times \{y \} \cap G_f) \neq \emptyset$ then $\displaystyle x \in X$ and $\displaystyle y = y$ so that $\displaystyle X \times \{y \} \cap G_f \neq \emptyset$. Then $\displaystyle \forall y \in Y, \exists x \in X, y = f(x)$ so that $\displaystyle f$ is surjective. $\displaystyle \square$

Is this proof correct?

Thanks

2. Originally Posted by tukeywilliams
Is this proof correct?
Yes it is correct and well done.