This is part of Exercise I.4 of "Sheaves in Geometry and Logic [. . .]" by Mac Lane and Moerdijk.
The Question:
Exercise: Let $F: \mathbf{A}\to \mathbf{B}$ be an equivalence of categories. Prove that a subobject classifier for $\mathbf{A}$ yields one for $\mathbf{B}$.
The Details:
Here equivalence of categories is defined as follows.
Definition: A functor $F: \mathbf{A}\to \mathbf{B}$ is an equivalence of categories if for any $\mathbf{A}$-objects $A, A'$, we have that
$$\begin{align}
{\rm Hom}_{\mathbf{A}}(A, A')&\to{\rm Hom}_{\mathbf{B}}(FA, FA')\\
p&\mapsto F(p)
\end{align}$$is a bijection and, moreover, any object of $\mathbf{B}$ is isomorphic to an object in the image of $F$.
A definition of a subobject classifier is given on page 32, ibid.
Definition: In a category $\mathbf{C}$ with finite limits, a subobject classifier is a monic, ${\rm true}:1\to\Omega$, such that to every monic $S\rightarrowtail X$ in $\mathbf{C}$ there is a unique arrow $\phi$ which, with the given monic, forms a pullback square
$$\begin{array}{ccc}
S & \to & 1 \\
\downarrow & \, & \downarrow {\rm true}\\
X & \stackrel{\dashrightarrow}{\phi} & \Omega.
\end{array}$$
My Attempt:
Let $F: \mathbf{A}\to \mathbf{B}$ be an equivalence of categories. Suppose ${\rm true}_{\mathbf{A}}:1_{\mathbf{A}}\to\Omega_{\mathbf{A}}$ is a subobject classifier of $\mathbf{A}$. Then there exists a functor $G:\mathbf{B}\to \mathbf{A}$ such that there are natural transformations $\alpha: F\circ G\stackrel{\sim}{\to}{\rm id}_{\mathbf{B}}$ and $\beta: G\circ F\stackrel{\sim}{\to}{\rm id}_{\mathbf{A}}$.
Showing $F$ of a terminal object is a terminal object . . .
Consider $Y\in{\rm Ob}(\mathbf{B})$ and $F(1_{\mathbf{A}})$. We have some $Y_{\mathbf{A}}\in{\rm Ob}(\mathbf{A})$ such that $F(Y_{\mathbf{A}})=Y$ (I'm not sure how to justify that) and
$$\begin{align}
{\rm Hom}_{\mathbf{A}}(Y_{\mathbf{A}}, 1_{\mathbf{A}})&\to{\rm Hom}_{\mathbf{B}}(Y, F(1_{\mathbf{A}}))\\
p&\mapsto F(p)
\end{align}$$
is a bijection. But $\lvert {\rm Hom}_{\mathbf{A}}(Y_{\mathbf{A}}, 1_{\mathbf{A}})\rvert=1$ as $1_{\mathbf{A}}$ is terminal. Hence $\lvert {\rm Hom}_{\mathbf{B}}(Y, F(1_{\mathbf{A}})\rvert=1.$ But $Y$ was arbitrary. Thus $F(1_{\mathbf{A}})$ is terminal in $\mathbf{B}$.
Showing $F$ of a monic is monic . . .
For any $M\stackrel{m}{\rightarrowtail} N$ monic in $\mathbf{A}$, for any $$L\overset{p}{\underset{q}{\rightrightarrows}}M\stackrel{m}{\rightarrowtail} N$$ such that if $m\circ p=m\circ q$, then $p=q$.
So $$[{\rm id}_{\mathbf{A}}(m\circ p)={\rm id}_{\mathbf{A}}(m\circ q)]\Rightarrow {\rm id}_{\mathbf{A}}(p)={\rm id}_{\mathbf{A}}(q);$$
that is,
$$[(G\circ F)(m\circ p)=(G\circ F)(m\circ q)]\Rightarrow (G\circ F)(p)=(G\circ F)(q);$$
then,
$$[F\circ(G\circ F)(m\circ p)=F\circ(G\circ F)(m\circ q)]\Rightarrow F\circ (G\circ F)(p)=F\circ (G\circ F)(q);$$
that is,
$$[(F\circ G)(F(m)\circ F(p))=(F\circ G)(F(m)\circ F(q))]\Rightarrow (F\circ G)(F(p))=(F\circ G)(F(q)).$$
But $F\circ G={\rm id}_{\mathbf{B}}$, so we have
$$[F(m)\circ F(p)=F(m)\circ F(q)]\Rightarrow F(p)=F(q),$$
so $F(M)\stackrel{F(m)}{\rightarrow}F(N)$ is monic in $\mathbf{B}$.
I'm not sure how to continue from here. It seems like the rest is diagram chasing but I want to understand it better than that.
Please help 🙂
Best Answer
Since you've asked for an answer using only the concepts in the question, I'll do my best to give such a proof. However, before I do, I want to say that Hanno's suggestion in the comments is definitely the best way to prove this.
Also your proofs have some issues. I won't point out all the differences, because this answer will be long enough as is, but please take note of them.
Proof:
As you've noted, we should prove that $F1$ is a terminal object of $D$, and that $F\alpha$ remains a monomorphism.
$F1$ is terminal: Let $X\in D$, since $F$ is essentially surjective, there exists $A\in C$ such that $FA\cong X$. Then $$D(X,F1)\cong D(FA,F1)\cong C(A,1)\cong \{*\}.$$ Thus $F1$ is terminal in $D$, since all hom sets $D(X,F1)$ are one element sets.
$F$ preserves (and reflects) monomorphisms: Let's rephrase the property of being a monomorphism. $i:X\to Y$ is a monomorphism in a category $C$ if for all objects $Z\in C$, $g,h:Z\to X$, $ig=ih \implies g=h$. In other words, the map $i_* : C(Z,X)\to C(Z,Y)$ is injective for all objects $Z\in C$.
Now suppose $i:X\to Y$ is a monomorphism in $C$, so $i_*:C(Z,X)\to C(Z,Y)$ is injective. Then we have the diagram $$\require{AMScd} \begin{CD} C(Z,X) @>i\circ ->> C(Z,Y)\\ @V\simeq VFV @V\simeq VFV \\ D(FZ,FX) @>Fi\circ - >> D(FZ,FY), \\ \end{CD} $$ so $Fi_*$ is injective a map from $D(FZ,FX)$ to $D(FZ,FY)$ for all $Z\in C$. This is not quite what we want though, but it's good enough. For any object $A\in D$, we can find $Z\in C$ such that $\phi : A\xrightarrow{\cong} FZ$. Then the composite $$D(A,FX)\newcommand\toby\xrightarrow \toby{-\circ \phi^{-1}} D(FZ,FX) \overset{Fi\circ}{\hookrightarrow} D(FZ, FY) \toby{-\circ\phi} D(A,FY)$$ is injective, and reduces to just the map $f\mapsto Fi\circ f \circ \phi^{-1}\circ \phi = Fi\circ f$. Thus $Fi$ is a monomorphism in $D$.
In particular, $F\alpha : F1\to F\Omega$ is a monomorphism.
Note also that if $Fi$ is a monomorphism in $D$, then by the commutative square above, we see that $i$ must have been a monomorphism in $C$.
$F\alpha : F1\to F\Omega$ is a subobject classifier: Let $i : A\to X$ be a monomorphism in $D$. Choose $B,Y\in C$ such that $A\cong FB$ and $X\cong FY$. Let $J : FB\to FY$ be the conjugate monomorphism of $i$. Fixing these isomorphisms, then there is a unique morphism $\varphi : X\to F\Omega$ making $$ \begin{CD} A @>>> F1 \\ @VVi V @VV F\alpha V\\ X @>\varphi>> F\Omega \\ \end{CD} $$ a pullback square if and only if there is a unique morphism $\Phi : FY\to F\Omega$ making $$ \begin{CD} FB @>>> F1 \\ @VVJ V @VV F\alpha V\\ FY @>\Phi>> F\Omega \\ \end{CD} $$ a pullback square. Thus it suffices to prove that there is such a unique morphism $\Phi$.
Now since $F:C(B,Y)\to D(FB,FY)$ is a bijection, there is some morphism $j:B\to Y$ such that $J=Fj$. By the note in the previous section, $j$ is a monomorphism. Thus since $\alpha$ is a subobject classifier in $C$, we have a unique morphism $\psi:Y\to \Omega$ such that the following diagram is a pullback square: $$ \begin{CD} B @>>> 1 \\ @VVj V @VV \alpha V\\ Y @>\psi>> \Omega. \\ \end{CD} $$
It just remains to prove that $\Phi = F\psi$ has the desired property. Certainly the square commutes, and if $\Phi$ were a morphism making the square commute, then since $F$ induces a bijection on hom sets, we could find a map $\psi$ such that $F\psi=\Phi$ making the corresponding square in $C$ commute. Thus it suffices to prove that $$ \begin{CD} B @>>> 1 \\ @VVj V @VV \alpha V\\ Y @>\psi>> \Omega \\ \end{CD} $$ is a pullback square if and only if $$ \begin{CD} FB @>>> F1 \\ @VVFj V @VV F\alpha V\\ FY @>F\psi >> F\Omega \\ \end{CD} $$ is a pullback square.
This is the same idea as before when we were proving $F1$ was terminal, or that $Fi$ is a monomorphism if and only if $i$ is a monomorphism. In fact, the general fact is that if $F$ is an equivalence of categories, then it both preserves and reflects all limits and colimits. I'll prove that $F$ preserves and reflects all limits, since the general language of cones to a diagram is more convenient than talking about pullbacks specifically.
$F$ preserves and reflects all limits:
Proof
First suppose $(FX,\set{Ff_a})$ is a limit cone. Then given a cone $(Y,\set{g_a})$ to $I$, the cone $(FY,\set{Fg_a})$ induces a unique map $\alpha_0 : FY\to FX$ such that $Ff_a \circ \alpha_0 = Fg_a$. Let $\alpha : Y\to X$ be the corresponding map, so that $F\alpha = F\alpha_0$. Then we have that $f_a \circ \alpha = g_a$, and $\alpha$ is unique, since if $\tilde{\alpha}$ also had this property, then $F\tilde{\alpha} : FY\to FX$ would satisfy the same property as $\alpha_0$. So by uniqueness, $F\tilde{\alpha} = \alpha_0= F\alpha$. Hence $\tilde{\alpha}=\alpha$. Thus $(X,\set{f_a})$ is a limit cone. Therefore $F$ reflects limits.
Now suppose $(X,\set{f_a})$ were a limit cone. If $(Z,\set{G_a})$ is a cone to $FI$ in $D$, then choose $Y\in C$ with $Z\cong FY$. Then we have a corresponding cone $(FY,\set{G_a'})$. As usual, we can now reflect the morphisms into $C$ to get $g_a : Y\to Ia$ such that $(Y,\set{g_a})$ is a cone to $I$ in $C$. Then the unique morphism $Y\to X$ induced by this cone becomes a unique morphism $FY\to FX$, and thus a unique morphism $Z\to FX$. Thus $(FX,\set{Ff_a})$ is a limit cone in $D$.
This completes the proof of both the lemma, and the original claim. $\blacksquare$