This is the second part of Exercise I.4 of Mac Lane & Moerdijk's, "Sheaves in Geometry and Logic [. . .]".
The Details:
First we have
Definition 1: 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 $F$.
From p. 17 ibid. . . .
Definition 2: Given two functors
$$F:\mathbf{X}\to \mathbf{A}\quad G: \mathbf{A}\to \mathbf{X},$$
we say that $G$ is right adjoint to $F$, written $F\dashv G$, when for any $X\in{\rm Ob}(\mathbf{X})$ and any $A\in{\rm Ob}(\mathbf{A})$, there is a natural bijection between morphisms
$$\frac{X\stackrel{f}{\to}G(A)}{F(X)\stackrel{h}{\to}A},$$
in the sense that each $f$, as displayed, uniquely determines $h$, and conversely.
There is more to this definition on page 18.
From p. 19 ibid. . . .
Definition 3: Suppose products exist in $\mathbf{C}$. For a fixed $A\in{\rm Ob}(\mathbf{C})$, one may consider the functor
$$A\times -: \mathbf{C}\to \mathbf{C}.$$
If this functor had a right adjoint (unique up to isomorphism), this adjoint is denoted by
$$(-)^A:\mathbf{C}\to \mathbf{C}.$$
In this case $A$ is said to be an exponentiable object of $\mathbf{C}$.
From p. 20 ibid. . . .
Definition 4: A category $\mathbf{C}$ is a cartesian closed category (CCC) if
it has all finite products (which is equivalent to saying there exists a terminal object and all binary products in $\mathbf{C}$) and
all $\mathbf{C}$-objects are exponentiable.
The Question:
Let $F:\mathbf{A}\to\mathbf{B}$ be an equivalence of categories. Suppose $\mathbf{A}$ is a CCC. Show that $\mathbf{B}$ is a CCC.
Context:
I am teaching myself topos theory for fun.
I have read (but not fully understood) Goldblatt's, Topoi: [. . .].
Previous questions along these lines from myself include the following:
But these are from several years ago.
My Attempt:
Terminal object . . .
From the reasoning in
I may conclude that $\mathbf{B}$ has a terminal object.
Binary products . . .
Suppose that $Y, B, B'\in{\rm Ob}(\mathbf{B})$. Then $Y=F(X), B=F(A), B'=F(A')$ for some $X, A, A'\in{\rm Ob}(\mathbf{A})$.
Since $\mathbf{A}$ is a CCC, the product $A\times A'$ exists. So there exist morphisms $\pi_1: A\times A'\to A$ and $\pi_2: A\times A'\to A'$ such that for any $p_1: X\to A$ and any $p_2: X\to A'$, there exists a unique $u: X\to A\times A'$ such that
$$p_1=\pi_1\circ u\quad\text{and}\quad p_2=\pi_2\circ u.$$
Consider $\widetilde{p_i}=F(p_i)$.
We have, by the equivalence, that
$$
{\rm Hom}_{\mathbf{A}}(X, A\times A')\cong {\rm Hom}_{\mathbf{B}}(Y, F(A\times A')),$$
but $\lvert{\rm Hom}_{\mathbf{A}}(X, A\times A')\rvert=\lvert\{u\}\rvert=1,$ so $F(u): Y\to F(A\times A')$ is unique.
Also by equivalence 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 $\beta: G\circ F\stackrel{\sim}{\to}{\rm id}_{\mathbf{A}}$.
Define $\widetilde{\pi_i}=F(\pi_i)$. Then we have
$$\begin{align}
\widetilde{\pi_i}\circ F(u)=\widetilde{p_i}&\iff F(\pi_i)\circ F(u)=F(p_i) \\
&\iff F(\pi_i\circ u)=F(p_i) \\
&\iff (G\circ F)(\pi_i\circ u)=(G\circ F)(p_i) \\
&\iff {\rm id}_{\mathbf{A}}(\pi_i\circ u)={\rm id}_{\mathbf{A}}(p_i) \\
&\iff \pi_i\circ u=p_i,
\end{align}$$
which holds by definition of $u$ (and $G$ is as defined above).
Thus $B\times B'=F(A\times A')$ exists.
Exponents . . .
Since $\mathbf{A}$ is a CCC, each $A, A'\in {\rm Ob}(\mathbf{A})$ is exponentiable.
From $A\times -\dashv (-)^A$, we have
$$\frac{A'\stackrel{f}{\to}A\times A'}{A'^A\stackrel{g}{\to}A'}$$
for some $f, g\in {\rm Mor}(\mathbf{A})$. But then
$$\frac{F(A'\stackrel{f}{\to}A\times A')}{F(A'^A\stackrel{g}{\to}A')},$$
i.e.,
$$\frac{F(A')\stackrel{F(f)}{\to}F(A\times A')=B\times B'}{F(A'^A)\stackrel{F(g)}{\to}F(A')=B'}.$$
But here I am stuck. What does $F(A'^A)$ correspond to in $\mathbf{B}$?
I'm guessing $F(A'^A)\stackrel{?}{\equiv}F(A')^{F(A)}\stackrel{?}{\equiv}B'^B$ in some sense . . .
Please help 🙂
Best Answer
An equivalence of categories $F$ it is handier with it's quasi-inverse. And since equivalences can be promoted to adjoint equivalences, let's just take that too.
So consider $G$ to be the right adjoint quasi-inverse to $F$ (since they are equivalences it is also a left adjoint). Now with this we have natural isomorphisms that will help. First of all the product in $\mathbf{B}$ can be written $F(G(B)\times G(B'))$. Consider the functor $$Y \mapsto \text{Hom}_{\mathbf{B}}(F(G(Y)\times G(B)),B').$$ We have the following chain of equivalences natural in $Y$, $$\begin{align} \text{Hom}_{\mathbf{B}}(F(G(Y)\times G(B)),B')&\simeq \text{Hom}_{\mathbf{A}}(G(Y)\times G(B),G(B'))\\ &\simeq \text{Hom}_{\mathbf{A}}(G(Y),G(B')^{G(B)})\\ &\simeq \text{Hom}_{\mathbf{B}}(Y,F(G(B')^{G(B)})). \end{align}$$ This show that $F(G(B')^{G(B)})$ is the exponent object.