# 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.

• What is your definition of "subobject classifier"? – Eric Wofsey Nov 3 '18 at 20:27
• I would have thought that "subobject classifier is the representing object of the subobject functor" was the definition.... – Angina Seng Nov 3 '18 at 20:31
• Hint: Use that isomorphism of any $C$ with the object part of a pullback induces a pullback structure on $C$. – Malice Vidrine Nov 3 '18 at 20:35
• Edited to give the definition – Adi Ostrov Nov 3 '18 at 20:38

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 $$\chi_m$$.