$\def\Rng{\operatorname{Rng}}$ Let $R$ be a relation from $A$ to $B$. For $a \in A$, define the vertical section of $R$ at $a$ to be $R_a$ = $\{ y \in B: (a,y) \in R\}$. Prove that the union over $R_a$ where $a \in A = \Rng(R)$.
I have work for it but as you can see I have no formatting abilities, sorry! I worked through the definitions of $R_a$ and $\Rng(R)$ and came to the proper conclusions by showing inclusion in both directions but it seems iffy to me.
So if $y \in \bigcup_{a \in A} R_a \Rightarrow y \in R_x$ for some $x \in A$, then $(x, y) \in R$, hence $y \in \operatorname{Rng}(R)$. Conversely $y \in \operatorname{Rng}(R)$ implies that there exists $x \in A$, such that $(x, y) \in R \Rightarrow y \in R_x \subseteq \bigcup_{a\in A} R_a$.