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 $C$.
Best Answer
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 $\mathscr{C}$.