# Subobject classifier of $C$ is the same as the terminal object of $D$

Let $$C$$ be a category, then construct the category $$D$$ whose objects are monomorphisma of $$C$$ and arrows are pullback squares of these monics. Show that the subobject classifier of $$C$$ is the same as the terminal object of $$D$$.

I tried this: Let $$1\rightarrow \Omega$$ be the subobject classifier of $$C$$. Then for each monic arrow $$a\rightarrow b$$, I have to show there is a diagram

$$\require{AMScd}$$ $$\begin{CD} a @>{f}>> b\\ @VVV @VVV\\ 1 @>{g}>> \Omega \end{CD}$$ Which is a pullback...

$$a\rightarrow b$$ and $$1\rightarrow \Omega$$ are monic by definition.... so now I have no idea what to do.... do I have to show $$b\rightarrow \Omega$$ is monic and use the fact that monics are stable under pull back?

Now let $$c\rightarrow d$$ be the terminal object in $$D$$, then $$c\rightarrow d$$ is monic, then for a monic $$t\rightarrow k$$ there are maps $$t\rightarrow c$$ , $$k\rightarrow d$$ such that the diagram

$$\require{AMScd}$$ $$\begin{CD} t @>{i}>> c\\ @VVV @VVV\\ k @>{j}>> d \end{CD}$$ Is a pullback. Then $$c\rightarrow d$$ is a subobject classifier of $$C$$.

• Well, what is the universal property of the subobject classifier? May 5 at 21:48
• I think it is the same as the universal property of fibered products.... May 5 at 21:50
• Double-check the universal property of $\Omega$, because afaik it basically says exactly that for any monomorphism $a\to b$ you can find the pullback square you're looking for (and uniquely). May 5 at 21:53
• I think @shibai is right: If you look carefully at the definition of "subobject classifier in $C$" and the definition of "terminal object in $D$", you'll find that they say the same thing. May 6 at 1:42

Here's a quick proof that I was able to write up. I'll write your category $$C$$ as $$\mathscr{C}$$ and your category $$D$$ as $$\mathscr{D}$$ and then show that $$\operatorname{true}:\top \to \Omega$$ is a terminal object in $$\mathscr{D}$$ (note that $$\top$$ is the terminal object of $$\mathscr{C}$$). First let $$\mu:A \to B$$ be a monic in $$\mathscr{C}$$ and note that because $$\Omega$$ is a subobject classifier in $$\mathscr{C}$$ there is a unique morphism $$\chi_{\mu}:B \to \Omega$$ which makes the diagram $$\begin{array} S A & \xrightarrow{!_A} & \top \\ \mu\downarrow & & \downarrow \operatorname{true} \\ B & \xrightarrow[\chi_{\mu}]{} & \Omega \\ {} \end{array}$$ into a pullback diagram. This gives a morphism $$\Phi:\mu \to \operatorname{true}$$ in $$\mathscr{D}$$.
To complete our verification that this is a terminal object it suffices to show that the pullback square $$\Phi$$ is the unique such pullback square with vertical edges $$\mu$$ and $$\operatorname{true}$$. However, this is more or less immediate: Because $$\top$$ is the terminal object in $$\mathscr{C}$$ there is exactly one morphism from $$A$$ to $$\top$$; as such, if there are any other pullback squares in $$\mathscr{C}$$ with vertical edges $$\mu$$ and $$\operatorname{true}$$, they differ only via the bottom edge of the square. However, by the definition of the subobject classifier $$\Omega$$ the map $$\chi_{\mu}$$ is the unique morphism $$B \to \Omega$$ which renders the square as a pullback square, so we conclude that $$\Phi$$ must indeed be unique. Since we have a unique morphism from $$\mu$$ to $$\operatorname{true}$$, we conclude that $$\operatorname{true}$$ is a terminal object of $$\mathscr{D}$$.
Alright, here's the converse I promised. Let $$m:X \to Y$$ be a terminal object in $$\mathscr{D}$$. We claim that $$X$$ is a terminal object of $$\mathscr{C}$$. This can be seen by noting that for any object $$A$$ of $$\mathscr{C}$$, the identity morphism is monic so there is a unique pullback square $$\begin{array} S A & \xrightarrow{f} & X \\ \operatorname{id}_A\downarrow & & \downarrow m \\ A & \xrightarrow[g]{} & Y \\ {} \end{array}$$ in $$\mathscr{C}$$. This allows us to deduce that first every object $$A$$ has a map into the object $$X$$; this map is unique by the uniqueness of the pullback square (which follows from the fact that $$m$$ is terminal), so $$X$$ is indeed a terminal object of $$\mathscr{C}$$. Finally, when given an arbitrary monic $$\mu:A \to B$$ in $$\mathscr{C}$$, consider the unique pullback square $$\begin{array} S A & \xrightarrow{f} & X \\ \mu\downarrow & & \downarrow \operatorname{m} \\ B & \xrightarrow[g]{} & Y \\ {} \end{array}$$ in $$\mathscr{C}$$. We already know that because $$X$$ is a terminal object, the top edge of the diagram is the unique morphism $$A \to X$$. Using the uniqueness of the pullback allows us to deduce that $$g$$ is the unique map $$B \to Y$$ making the square a pullback square and hence realized $$g$$ as the classifying map of $$\mu$$. Thus $$Y$$ is a subobject classifier in $$\mathscr{C}$$.