As a simple exercise, I wanted to write up a proof that in any topos $\mathcal{E}$, given a subobject $r:R\rightarrowtail X\times Y$ in $\mathcal{E}$, if $$(\forall x:X)(\forall y,z:Y)([R(x,y)\wedge R(x,z)]\Rightarrow y=z)$$ holds in the internal language, then the projection $\pi_1r:R\to X$ is a monomorphism. For some reason, I have been unable to see my way to a solution after a couple of days of picking at it, and I've started just going in circles.
Unless I'm forgetting how any of this works, the above sentence holding in the internal language ought to mean that $$\pi_{1,2}^*r\cap\pi_{1,3}^*r\leq \pi_{2,3}^*\Delta_Y\hspace{3em}(*)$$ in the subobject lattice of $X\times Y\times Y$ (where $\Delta_Y$ is the diagonal map $Y\to Y\times Y$). It's easy enough to show that if $(\pi_1 r\times\pi_1r)^*\Delta_Y\leq \Delta_R$ holds in the subobject lattice of $R\times R$ then $\pi_1r$ is a monomorphism, so my main strategy has been to try and derive this from $(*)$; this hasn't worked for me yet.
tl;dr - How can I show that $\pi_1r$ is monomorphic, starting from the assumption $(*)$?
You can do it directly : if $u,v$ are maps $Z\to R$ such that $\pi_1ru=\pi_1rv$, then to prove that $u=v$ it suffices to prove that $ru=rv$ (since $r$ is a mono), and for that is enough to prove that $\pi_2ru=\pi_2rv$, since $\pi_{1}$ and $\pi_2$ are jointly monic. Since the condition $(*)$ tells you that $\pi_1ru=\pi_1rv\Rightarrow \pi_2ru=\pi_2rv$, you're done !