Q:

Show that $A$ cartesian closed need not imply $A^J$ is cartesian closed. 

I need to find an example of a cartesian closed category whose functor category (from some diagram $J$) is not cartesian closed. I thought of one possible solution: let $A$ be a commutative square (which is a poset, hence ccc) and let $J$ be the category $\bullet \to \bullet$. Then $A^J$ is the arrow category, and I claim it's impossible to form the product of the parallel arrows in $A$'s square. This seems like a poor example though, any other suggestions?

A:

I think it will not be easy to find an example with $J=2$.  Since a cartesian closed category has finite products, the result cited in David White's answer shows the category $A$ for such an example must not have equalizers.  So it cannot be a preorder, and so (having binary products) it cannot be finite.  It is not hard to find infinite cartesian closed categories without equalizers -- but I have not found one with a specific pair of functors that I can show have no exponential.
The best example that occurs to me has $J$ the poset with bottom element $0$ and a countable infinity of objects right above $0$, with no arrows to each other.  That is the partial order on the natural numbers with $x\leq y$ if and only if $x=0$.  And for $A$ the category of finite sets (hereditarily finite, if you like ZF foundations and want a small category).  The result follows since the functor assigning the empty set to $0$ and the two element set to every other object of $J$ has infinitely many natural transformations to itself (uncountably many).

