The composition $\mathcal{R}_{2}\circ\mathcal{R}_{1}$ of the relations $\mathcal{R}_{1}$ e $\mathcal{R}_{2}$ is defined as follows:
$$\mathcal{R}_{2}\circ\mathcal{R}_{1} := \{(x,z) \mid \exists y\,(x\mathcal{R}_{1}y\wedge y\mathcal{R}_{2}z)\}$$
In particular, if $\mathcal{R}_{1}\subset X\times Y$ and $\mathcal{R}_{2}\subset Y\times Z$, then $\mathcal{R} = \mathcal{R}_{2}\circ\mathcal{R}_{1}\subset X\times Z$ and $$x\mathcal{R}z := \exists y\,((y\in Y)\wedge(x\mathcal{R}_{1}y)\wedge(y\mathcal{R}_{2}z))$$
Let $\Delta_{X}$ be the diagonal of $X^{2}$ and $\Delta_{Y}$ the diagonal of $Y^{2}$. Show that if the relations $\mathcal{R}_{1}\subset X\times Y$ and $\mathcal{R}_{2}\subset Y\times X$ are such that $(\mathcal{R}_{2}\circ\mathcal{R}_{1} = \Delta_{X})\wedge(\mathcal{R}_{1}\circ\mathcal{R}_{2} = \Delta_{Y})$, then both relations are functional and define mutually inverse mappings of $X$ and $Y$.
MY ATTEMPT
I know that for a relation to be functional such relation has to satisfy \begin{align*} (x\mathcal{R}y)\wedge(x\mathcal{R}z) \Longrightarrow y = z \end{align*}
Since $\mathcal{R}_{2}\circ\mathcal{R}_{1} = \Delta_{X}$, for any pair $(x,x)\in\Delta_{X}$, one has \begin{align*} (x\mathcal{R}_{2}\circ\mathcal{R}_{1}x) \Longrightarrow\exists y\,((y\in Y)\wedge(x\mathcal{R}_{1}y)\wedge(y\mathcal{R}_{2}x)) \end{align*}
Analogously, since $\mathcal{R}_{1}\circ\mathcal{R}_{2} = \Delta_{Y}$, for any pair $(y,y)\in\Delta_{Y}$, one has \begin{align*} (y\mathcal{R}_{1}\circ\mathcal{R}_{2}y) \Longrightarrow\exists x\,((x\in X)\wedge(y\mathcal{R}_{1}x)\wedge(x\mathcal{R}_{2}y)) \end{align*}
Then I get stuck. Could someone help me out?
Your work shows that the domain of both relations is full, and if they are functional, then $R_2\circ R_1(x)=x$ and $R_1\circ R_2(y)=y$ for all $x\in X $ and $y\in Y$, so they are inverses to each other.
Thus, only functionality is left to prove.
Let $xR_1y$ and $xR_1z$. Then $\exists x': yR_2x'R_1y$, but as $x(R_2\circ R_1)x'$, we must have $x=x'$ by hypothesis.
Now this implies $yR_2xR_1z$, so $y(R_1\circ R_2)z$ which yields $y=z$ again by the hypothesis.