# 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?

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Why do you need this example? –  Steven Landsburg Jul 16 '12 at 17:14
I'm working on an answer key for Categories for the Working Mathematician and got stuck on this problem (IV.6 #5) –  Caleb Weinreb Jul 16 '12 at 17:43
@Caleb, That sounds like a nice project. Will you have notes? –  Spice the Bird Aug 16 '12 at 4:27
Here is a link to the work in progress... It covers most exercises from chapter I - VI scribd.com/doc/103093298/Mac-Lane-Answers –  Caleb Weinreb Aug 17 '12 at 1:42

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).

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Nice comment about choice of foundations! –  David Roberts Aug 16 '12 at 6:36
Welcome to MO ! –  Steven Gubkin Aug 16 '12 at 16:28
Sorry, can I you elaborate on why the functor you described makes this setup a counterexample? I don't understand. Thanks! –  Caleb Weinreb Aug 17 '12 at 2:29
Since $J$ has an initial object $0$, for any $\mathcal{F}:J\rightarrow A$ the value $\mathcal{F}(0)$ is the set of natural transformations $1\rightarrow \mathscr{F}$ where $1$ is the constant functor with singleton set as value. So if functors $\mathcal{F,G}:J\rightarrow A$ have exponential $\mathscr{G^F}$ in the functor category, then $\mathscr{G^F}(0)$ is the set of natural transformations $\mathscr{F}\rightarrow\mathscr{G}$. So the exponential cannot exist for functors with infinitely many natural transformations between them. This functor has infinitely many to itself. –  Colin McLarty Aug 17 '12 at 15:18

Your example works, but only if $A$ is badly behaved. (EDIT: As Colin points out in his comment, the OP's idea was to take $A$ as a cartesian closed poset, and that won't work. However, keeping $J$ as $\bullet \to \bullet$ will yield an example where $A$ is cartesian closed but $A^J$ is not). In particular, this link gives the following (paraphrased) equivalence:

For $J = \bullet \rightarrow \bullet$, $A^J = Arr(A)$ is cartesian closed if and only if $A$ is cartesian closed and, for all arrows $f: a \rightarrow b$ and $g: c \rightarrow d$ in $A$, the arrows $[f,1]: [b,d] \rightarrow [a,d]$ and $[1,g] : [a,c] \rightarrow [a,d]$ admit a pullback $P(f,g)$ in $A$.

Here $[a,d] = d^a = Hom(a,d)$ exists because $A$ is cartesian closed. That link also says this result is a special case of the considerations in the following paper, since the category $A^J$ is obtained from $A$ by Artin gluing.

Aurelio Carboni, Peter Johnstone, Connected limits, familial representability and Artin glueing, Mathematical Structures in Computer Science 5 (1995), 441--459

If you need a more explicit example that $A^J$ can fail to be closed, it might be found there. Note that in all applications I've ever used, this extra hypothesis about pullbacks is true and so $A^J$ has two closed cartesian structures. In the first, the product is

$f\otimes g: a \otimes c \rightarrow b\otimes d$

and the closed structure is given by the projection

$[a,c] \times_{[a,d]} [b,d] \rightarrow [b,d]$

In the second, the product is

$f\Box g: a \otimes d \coprod_{a\otimes c} b\otimes c \rightarrow b\otimes d$

and the closed structure is given by the map induced by the pullback

$[b,c] \rightarrow [a,c]\times_{[a,d]} [b,d]$

You see where the pullbacks are necessary. Of course, the terminal object is simply $1_\ast$ where $\ast$ is the terminal object in $A$.

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Thank you so much! –  Caleb Weinreb Jul 25 '12 at 1:51
The result you link to actually shows that nothing like this purported example can work. Precisely: it shows for any cartesian closed poset $A$, the arrow category $A^2$ is cartesian closed, because a poset with (all binary) products always has (all binary) pullbacks. And that cited result is easily proved directly from the definitions. –  Colin McLarty Aug 14 '12 at 15:46
@Colin McLarty: I guess I missed the bit where the OP asked that $A$ be a poset. In my answer I was thinking of $A$ as a general cartesian closed category. If such $A$ fails the conditions on $[f,1]$ and $[1,g]$ then $A^J$ is not cartesian closed, and the OP has the example he wants. Thanks for pointing out this bit about posets. I'd hate for someone to come upon this later and be confused by that. –  David White Aug 14 '12 at 21:44