Exercise 4 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: % 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.