Theorem: An equivalence of categories preserves a subobject classifier

Let $\mathbf{C}$ and $\mathbf{D}$ be equivalent, with $\Omega_{\mathbf{C}}$ a subobject classifier in $\mathbf{C}$. Let $F : \mathbf{C} \rightarrow \mathbf{D}$ and $G : \mathbf{D} \rightarrow \mathbf{C}$ be the equivalence. We may choose $F$ and $G$ to be an adjoint equivalence, so that $F \dashv G$, and $\eta : 1 \rightarrow GF$ and $\varepsilon : FG \rightarrow 1$ are natural isomorphisms.

What we wish to show is that $\text{Sub}_{\mathbf{D}}(\dash) \cong \Hom(\dash, F(\Omega_{\mathbf{C}}))$ -- this means precisely that $F(\Omega_{\mathbf{C}})$ is a subobject classifier in $\mathbf{D}$ ($F(\Omega_{\mathbf{D}})$ is a representing object for the subobject functor). But since $F$ and $G$ are adjoint, the right-hand functor is isomorphic to $\Hom(G(\dash), \Omega_{\mathbf{C}}) = \text{Sub}_{\mathbf{C}}(G(\dash))$. So, it is sufficient to show that $\text{Sub}_{\mathbf{C}}(G(\dash)) \cong \text{Sub}_{\mathbf{D}}(\dash)$.

Indeed, we will define a natural transformation $\alpha : \text{Sub}_{\mathbf{C}}(G(\dash)) \rightarrow \text{Sub}_{\mathbf{D}}(\dash)$ as $\alpha_X = \varepsilon_X \circ F(\dash)$. Notice that this assigns to an arrow $i : S \rightarrow G(X)$ precisely its transpose $\overline{i}$ under the adjunction $F \dashv G$. This shows that each $\alpha_X$ is an isomorphism (it is precisely the homset bijection of the adjunction), so we need only show that the transformation is natural. Keeping in mind that $\text{Sub}$ operates on arrows "by pullback", we have the naturality diagram for $f : X \rightarrow Y$:

Take a subobject $i : S \xrightarrow{} G(Y)$ at the top-right. Following the top left path, we pullback along $G(f)$ to obtain $P \xrightarrow{} G(X)$, and then take the transpose to get $F(P) \xrightarrow{} FG(X) \xrightarrow{\sim} X$. Following the other direction, we take the transpose to obtain $F(S) \xrightarrow{} FG(Y) \xrightarrow{\sim} Y$, and then we must pull this back along $f$. To say that that these are the same amounts to saying that the outer rectangle of this combined diagram is a pullback:

Indeed, the top square is a pullback because it is the image under $F$ of a pullback, and viewing $F$ as a right adjoint (after inverting the unit and counit) means it preserves limits. The bottom square can be shown a pullback directly by considering another cone $X \xleftarrow{p_2} P \xrightarrow{p_1} FG(Y)$. Simply define $\phi = \varepsilon_X^{-1} \circ p_2$. We have $p_2 = \varepsilon_X \circ \phi$ immediately. For the top:

$$\begin{align*} & \varepsilon_Y \circ p_1 = f \circ p_2 && \text{From the new cone} \\ \Rightarrow \enspace & \varepsilon_Y \circ p_1 = f \circ \varepsilon_X \circ \phi && \text{Substitute for } p_2 \\ \Rightarrow \enspace & \varepsilon_Y \circ p_1 = \varepsilon_Y \circ FG(f) \circ \phi && \text{From the original cone} \\ \Rightarrow \enspace & p_1 = FG(f) \circ \phi \end{align*}$$

Uniqueness of $\phi$ is immediate from the definition.

The pasting lemma then confirms that the outer rectangle is a pullback, and thus we have shown that the $\alpha$ is natural and thus a natural isomorphism, completing the proof.