Lifting subobject classifier of a category to its arrow category I'm reading Goldblatt's Topoi and trying to practice categorical reasoning, generalizing the example of $\mathbf{Set}^\rightarrow$ being a topos.
So, let $\mathcal{C}$ be a category with a subobject classifier $\Omega, \top$ where $\top : \mathbf{1} \rightarrow \Omega$. Consider the arrow category $\mathcal{C}^\rightarrow$: does it have a subobject classifier?
I think that it does: $\text{id}_\Omega, (\top, \top)$ seems like a good candidate. In particular, if $f, g$ are some monic arrows in $\mathcal{C}$ with characters $\chi_f, \chi_g$, then $(\chi_f, \chi_g)$ is the character of $(f, g)$ in $\mathcal{C}^\rightarrow$.
The construction I'm having in mind seems to follow from the structure of the proof I've done earlier that if $\mathcal{C}$ has pullbacks, then so does $\mathcal{C}^\rightarrow$ (namely, "gluing" two pullbacks of $\mathcal{C}$ in a cube-like diagram naturally compatible with the arrow category structure produces a pullback in $\mathcal{C}^\rightarrow$), but said proof is quite lengthy, so I'm omitting it here. The uniqueness of characters follows from the uniqueness of characters in $\mathcal{C}$.
So, my questions:


*

*If $\mathcal{C}$ has S.C.s, then does $\mathcal{C}^\rightarrow$ also have them?

*If it does, is the above indeed a S.C.?

*If it is, why Goldblatt is using a seemingly more involved construct for the specific example of $\mathbf{Set}^\rightarrow$?

 A: $\require{AMScd}$If $\cal E$ is an elementary topos, then so is the functor category ${\cal E}^C$ for every small category $C$; the classifying monomorphism is just the image of ${\sf true}:1 \hookrightarrow \Omega$ along the diagonal functor $\Delta : {\cal E} \to {\cal E}^C$. 
To see this, you can consider the obvious map ${\cal E}^C(G, \Delta\Omega) \to \text{Sub}(G)$ that sends a morphism $g : G \to \Delta\Omega$ to the pullback of the square
$$
\begin{CD}
S(g) @>>> \Delta1 \\
@Vm_gVV @VVV \\
G @>>g> \Delta\Omega
\end{CD}
$$
Limits and monics in functor categories are defined objectwise, so this just means taking the pullback objectwise; explicitly, for each  $x\in\cal E$
$$
\begin{CD}
S(g)x @>>> 1 \\
@Vm_{g,x}VV @VVV \\
Gx @>>g> \Omega
\end{CD}
$$ is a pullback in $\cal E$. Such pullback $S(g) \hookrightarrow G$ exists, because $\cal E$ is a topos, and it is a monic.
It remains to show that every monic arises in this way. Given such a monic $\alpha : F \Rightarrow G$, each $\alpha_x : Fx \to Gx$ in $\cal E$ corresponds to a certain $\chi_x^\alpha : Gx \to \Omega$ by virtue of the bijection $\text{Sub}(Gx)\cong {\cal E}(Gx,\Omega)$; since now the collection $\chi_x^\alpha : Gx \to \Omega$ forms a cocone for $G$, it corresponds to a unique  natural transformation $\bar\chi^\alpha : G \Rightarrow \Delta\Omega$. 
It is now just a matter of uwinding the definition to see that the pullback $m_{\bar\chi^\alpha} : S(\bar \chi^\alpha) \Rightarrow G$ coincides with $\alpha$, and conversely that if $g : G \Rightarrow \Delta\Omega$ is a morphism, $\bar\chi^{m_g} = g$.
A: The first question to ask would be: what are the monomorphisms of $\mathcal{C}^{\rightarrow}$?  To answer this question, let us use a Yoneda lemma style of argument: first, note that $\operatorname{Hom}_{\mathcal{C}^{\rightarrow}}(0 \to U, X \to Y) \simeq \operatorname{Hom}_{\mathcal{C}}(U, Y)$, and this isomorphism is natural in $Y$.  Thus, if $f : (X' \to Y') \to (X \to Y)$ is a monomorphism, then for every object $U$ of $\mathcal{C}$, we must have $f_2 \circ - : \operatorname{Hom}(U, Y') \to \operatorname{Hom}(U, Y)$ is injective; or in other words, $f_2 : Y' \to Y$ must be a monomorphism.  Similarly, $\operatorname{Hom}_{\mathcal{C}^\rightarrow}(\operatorname{id}_U : U \to U, X \to Y) \simeq \operatorname{Hom}_{\mathcal{C}}(U, X)$, and so we get that if $f : (X' \to Y') \to (X \to Y)$ is a monomorphism, then $f_1 : X' \to X$ must be a monomorphism.
I will now leave it as an exercise to show that $f_1 : X' \to X$ and $f_2 : Y' \to Y$ being monomorphisms in $\mathcal{C}$ is also sufficient to get $f : (X' \to Y') \to (X \to Y)$ is a monomorphism in $\mathcal{C}^{\rightarrow}$.

So, now suppose we have a subobject classifier $\Omega_1 \to \Omega_2$ in $\mathcal{C}^{\rightarrow}$.  Then let us focus first on $\Omega_2$: we will need that for any object $U$ of $\mathcal{C}$, we get
$$\operatorname{Hom}_{\mathcal{C}}(U, \Omega_2) \simeq \operatorname{Hom}_{\mathcal{C}^\rightarrow}(0\to U, \Omega_1 \to \Omega_2) \simeq \operatorname{Sub}_{\mathcal{C}^\rightarrow}(0 \to U).$$
Now by the above, if $(X' \to Y') \to (0 \to U)$ is a monomorphism, then $X' \to 0$ is a monomorphism, so $X'$ is an initial object also; and also, $Y' \to U$ is a monomorphism.  So, $\operatorname{Sub}_{\mathcal{C}^\rightarrow}(0 \to U) \simeq \operatorname{Sub}_{\mathcal{C}}(U)$; and all steps are natural in $U$.  Now, we ask: is there an object of $\mathcal{C}$ which represents the last functor $\operatorname{Sub}_{\mathcal{C}}$?  Yes: $\Omega_{\mathcal{C}}$ does by definition.
Similarly, we will need
$$\operatorname{Hom}_{\mathcal{C}}(U, \Omega_1) \simeq \operatorname{Hom}_{\mathcal{C}}(\operatorname{id}_U, \Omega_1 \to \Omega_2) \simeq \operatorname{Sub}_{\mathcal{C}^{\rightarrow}}(\operatorname{id}_U).$$
Now, to get a monomorphism $(X \to Y) \to \operatorname{id}_U$, we will need to have $X \to U$ and $Y \to U$ being monomorphisms.  In addition, we will need for $X\to U$ to factor through $Y \to U$ in order for there to be a map $X \to Y$ making a commutative diagram.  Thus,
\begin{align*}
\operatorname{Sub}_{\mathcal{C}^{\rightarrow}}(\operatorname{id}_U) & \simeq \{ (X, Y) \in \operatorname{Sub}_{\mathcal{C}}(U) \times \operatorname{Sub}_{\mathcal{C}}(U) \mid X \subseteq Y \} \\
& \simeq \{ (f, g) \in \operatorname{Hom}(U, \Omega) \times \operatorname{Hom}(U, \Omega) \mid f \le g \} \\
& \simeq \{ (f, g) \in \operatorname{Hom}(U, \Omega) \times \operatorname{Hom}(U, \Omega) \mid (f \rightarrow g) = \top \}.
\end{align*}
Again, all steps are natural in $U$; and we need to ask whether the last functor $\{ (f, g) : \operatorname{Hom}(-, \Omega) \times \operatorname{Hom}(-, \Omega) \mid (f \rightarrow g) = \top \}$ is representable.  I will now leave it as an exercise to show that the pullback in $\mathcal{C}$ of the morphisms $\rightarrow : \Omega_{\mathcal{C}} \times \Omega_{\mathcal{C}} \to \Omega_{\mathcal{C}}$ and $\top : 1 \to \Omega_{\mathcal{C}}$ does represent this functor.
Finally, the morphism $\Omega_1 \to \Omega_2$ will be induced by the morphism of functors $(0_U, \operatorname{id}_U) : (0 \to U) \to \operatorname{id}_U$; if you trace through the isomorphisms of functors, you will see that the corresponding morphism $\Omega_1 \to \Omega_2$ is the composition of the pullback map $\Omega_1 \hookrightarrow \Omega \times \Omega$ with $\pi_2 : \Omega \times \Omega \to \Omega$.

It now remains to see whether the "necessary values" of $\Omega_1 \to \Omega_2$ constructed above actually form a subobject classifier.  I will just comment that in order to get a subobject of $f : X \to Y$, you need a subobject $X' \hookrightarrow X$ and a subobject $Y' \hookrightarrow Y$.  Now, once you have those two things, you also need a map $X' \to Y'$ which makes the diagram commute; note that since $Y' \hookrightarrow Y$ is a monomorphism, there can be at most one such map.  Also note that such a map exists if and only if $X' \hookrightarrow X$ factors through the pullback $Y' \times_Y X \hookrightarrow X$.  I will leave it as an exercise to show that this is equivalent to the condition that $\chi_{X'} \le \chi_{Y'} \circ f$ as members of the Heyting algebra $\operatorname{Hom}_{\mathcal{C}} (X, \Omega)$; and then to use this to conclude that a subobject of $X \to Y$ is equivalent to a morphism $(X \to Y) \to (\Omega_1 \to \Omega_2)$ in $\mathcal{C}^{\rightarrow}$.
