Let $\bf E$ be an elementary topos, and let $t:1\to\Omega$ be its subobject classifier. Then, for every $x\in \mathrm{ob} (\mathbf E)$, there is a monic $\tau_x:u\to x\times \Omega^x$, universal in this sense: any monic $m:r\to x\times y$ is the pullback of $\tau_x$ along $1_x\times \rho$, for a unique $\rho:y\to \Omega^x$.
First, we need to define $\tau_x$, that I would set as the pullback of $t$ along the counit $\epsilon_\Omega:x\times\Omega^x\to\Omega$. Hence we obtain the following diagram, where the lower square is a cartesian. $\require{AMScd}$
$$\begin{CD} r @>{m}>> x\times y\\ @VV{\sigma}V @VV{1_x\times\rho}V \\ u @>{\tau_x}>> x\times\Omega^x \\ @VVV @VV{\epsilon_\Omega}V \\ 1 @>{t}>> \Omega \end{CD}$$
$\rho$ is defined as the arrow corresponding to $\chi_r:x\times y\to\Omega$ under the bijection $\Phi:\mathbf E(x\times y,\Omega)\to \mathbf E(y,\Omega^x)$, where $\chi_r$ is the characteristic map; $\sigma$ is the unique arrow (by the universal property of pullbacks) such that $\tau\sigma=(1_x\times \rho)m$. Actually, for the existence of $\sigma$ we need that the perimeter square is commutative; but it is sufficient that $\epsilon_{\Omega}(1_x\times\rho)=\chi_r$. This is true because $\epsilon_\Omega(1_x\times\rho)$ is exactly $\Phi^{-1}(\rho)=\chi_r$.
Thus by the pullback lemma, since the perimeter square and the lower square are cartesian, the upper one is too, that is our thesis. When $\mathbf E=\mathbf{Set}$, for example, $u$ is the subset of $x\times 2^x$ consisting of the pairs $(a,A)$ such that $a\in A$. Do you think my solution makes sense or do you have any correction? Thank you
Edit: I didn't prove that $\rho$ is unique, but if the upper square is cartesian, with $\rho'$ in place of $\rho$, also the perimeter square is cartesian; so $\Phi^{-1}(\rho')$ must be $\chi_r$, by the uniqueness hypothesis on the latter, showing $\rho'=\rho$.