Todd Trimble Cartesian closure of internal categories

We give a simple proof of the following result.

Theorem

Let $\mathbf{E}$ be a finitely complete cartesian closed category. Then the category $Cat(\mathbf{E})$ of internal categories in $E$ is also finitely complete and cartesian closed.

Proof

First suppose $\mathbf{E}$ is finitely complete. Then the category of directed graphs $\mathbf{E}^{\bullet \stackrel{\to}{\to} \bullet}$ is also finitely complete, and since $Cat(\mathbf{E})$ is monadic over $\mathbf{E}^{\bullet \stackrel{\to}{\to} \bullet}$, it follows that $Cat(\mathbf{E})$ is also finitely complete.

Now suppose that $\mathbf{E}$ is finitely complete and cartesian closed. Let $\Delta_3$ denote the category of nonempty ordinals up to and including the ordinal with 4 elements. We have a full and faithful embedding

$N \colon Cat(\mathbf{E}) \to \mathbf{E}^{\Delta_3^{op}}$

where the codomain category is cartesian closed. Indeed, the exponential of two objects $F$, $G$ in $\mathbf{E}^{\Delta_3^{op}}$ may be computed as an $\mathbf{E}$-enriched end

$G^F(m) = \int_n \prod_{f \colon n \to m} G(n)^{F(n)}$

when evaluated at $m \in Ob(\mathbf{E}^{\Delta_3^{op}})$, as is easily checked (see for instance here); note that this end is a finite limit diagram since $\Delta_3$ is finite.

If $C$, $D$ are internal categories in $\mathbf{E}$, seen as functors $\Delta_3^{op} \to \mathbf{E}$, the exponential $N C^{N D}$ defines the exponential in $Cat(\mathbf{E})$. To see this, it suffices to check that $N C^{N D}$, as defined by the end formula above, is a category $B$, i.e., is in the essential image of the nerve functor. For in that case, we have natural isomorphisms

$\array{ \arrayopts{\rowlines{solid}} F \times D \to C\;\;\; in\; Cat(\mathbf{E}) \\ N F \times N D \cong N(F \times D) \to N C\;\;\; in\; \mathbf{E}^{\Delta_3^{op}} \\ N F \to N C^{N D}\;\;\; in\; \mathbf{E}^{\Delta_3^{op}} \\ F \to B\;\;\; in \; Cat(\mathbf{E}) }$

whence $B$ satisfies the universal property required of an exponential.

Objects in the essential image of the nerve $N$ are characterized as functors $\Delta_3^{op} \to \mathbf{E}$ which take intervalic joins in $\Delta_3$ to pullbacks in $\mathbf{E}$, as given precisely by the Segal conditions. The remainder of the proof is then finished by the following lemma.

Lemma

If $C \colon \Delta_3^{op} \to \mathbf{E}$ satisfies the Segal conditions and $X \colon \Delta_3^{op} \to \mathbf{E}$ is any functor, then $C^X$ also satisfies the Segal conditions.

Proof

For any $X$ we have the formula

$C^X(m) = \int_k Hom(X k, \prod_{f \colon n \to k} \prod_{g \colon n \to m} C(n)).$

Since the enriched end and the internal hom-functor $Hom(X k, -)$ both preserve pullbacks, we are reduced to checking that

• If $C$ satisfies the Segal conditions, then so does
$\prod_{f \colon n \to k} \prod_{g \colon n \to m} C(n)$

as a functor $\Delta_3^{op} \to \mathbf{E}$ in the argument $m$ (for each fixed $k$).

Note that the displayed statement is a proposition in the language of finitely complete categories (i.e., in finitary essentially algebraic logic). Since hom-functors $\mathbf{E}(e, -) \colon \mathbf{E} \to Set$ jointly preserve and reflect the validity of such propositions, it suffices to prove it for the case where $\mathbf{E} = Set$. But this is classical elementary category theory; it says precisely that if $C$ is a small (ordinary) category, then the usual functor categories $C^{\mathbf{2}}$, $C^{\mathbf{3}}$ are equivalently described by exponentials of (truncated) simplicial sets. This completes the proof.

Revised on October 28, 2016 at 20:54:41 by Todd Trimble