I have been working through Goldblatt's Topoi and in Section 7.5 he proves the following result:
If $f,g$ and $h$ are monic with codomain $d$ in a topos, then $f\cap h\simeq g\cap h$ if and only if $\chi_f\circ h=\chi_g\circ h$.
This has led me to think about what the equalizer of $\chi_f$ and $\chi_g$ is, if it admits an easy characterization. It seems to me that it would be $f\cap g$, but I have not been able to prove it.
The arrow $f\cap g$ is equal to $g\circ g'=f\circ f'$ in the pullback of $f$ and $g$ (this is from Theorem 2 in Section 7.1): $\require{AMScd}$ \begin{CD} a\cap b @>g'>> b \\ @VVf'V @VVgV \\ a @>f>> d \end{CD}
So it is easy to see that it satisfies $\chi_f\circ f\cap g=\chi_g\circ f\cap g =\top_{a\cap b}$ if we just glue the pullback squares defining $\chi_f$ and $\chi_g$ and use the pullback lemma. Now, suppose that $h$ is monic into $d$ and $\chi_f\circ h=\chi_g\circ h$, i.e. $f\cap h \simeq g\cap h$ from the result in the beginning. Since ${\rm Sub} (d)$ is a lattice, then the isomorphism class $[f\cap h] \subseteq [f\cap g]$. Would this lead to $h\subseteq f\cap g$ , which would be the universal property of the equalizer? And even if it did, how could we prove this for non-monic $h$?
It will be the pullback of $\Delta:\Omega\to\Omega\times\Omega$ along $\langle \chi_f,\chi_g\rangle$, or, alternatively, the subobject classified by $=_\Omega\circ\langle\chi_f,\chi_g\rangle$. Since $=_\Omega:\Omega\times\Omega\to\Omega$ is the map corresponding to the logical operator $\Leftrightarrow$, one might call this $f\Leftrightarrow g$.
Thinking set-theoretically, this is like the set of all $x$ such that $x$ is in $f\cap g$ or it is in neither $f$ nor $g$. You can see this difference in terms of arrows because if we have $q:Y\to X$ such that $\chi_f\circ q=\bot!_Y$ and $\chi_g\circ q=\bot!_Y$, then $q$ will factor through $f\Leftrightarrow g$, but it will not factor through $f\cap g$.