Well Definedness of the Representation of the Subobject Functor Let $\mathcal{C}$ be a category with a subobject classifier $\Omega$.
I want to show that the suboject functor $Sub_{\mathcal{C}}(\cdot): \mathcal{C}^{op} \to \textbf{Set}$ is representable via $\Omega$. 
To do so I construct the natural transformation $\theta : Sub_\mathcal{C} \rightarrow \text{Hom}(\cdot,\Omega)$ given by $\theta_X([m]) = \chi_m$. with $[m]$ the equivalence class of $m$. I'd like to show that $\theta_X$ is well defined, that is if $m \cong m'$ as subobjects (there exists an isomorphism $\alpha : S \to S'$ with $m'\alpha = m$) then $\chi_m = \chi_{m'}$. 
I can't manage to show that. How can I do so?
EDIT:
I define that subobejct classifier in accordance with "Shaves in Geometry and Logic" where it is defined by saying that foreach monic $m$ there exists a unique $\chi_m$ such that the diagram is a pullback.

If $\chi_m$ is the classifying arrow of the monomorphism $m:S\to X$, and $m':S'\to X$ is another monomorphism with $\alpha:S\to S'$ an isomorphism, then $m\circ\alpha^{-1}=m'$ and $!_{S'}$ form a pullback diagram for $1\overset{\top}{\longrightarrow}\Omega\overset{\chi_m}{\longleftarrow}X$. By the universal property of the subobject classifier, there is exactly one arrow, $\chi_{m'}:X\to \Omega$, that can go in place of $\square$ make $\top\circ!_{S'}=\square\circ m'$ a pullback square. Since $\chi_m$ makes this a pullback, $\chi_{m'}$ must be identical with $\chi_m$.
