## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Discussion Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeNov 7th 2012

double categories are cartesian closed and it seems pretty reasonable that n-fold categories are cartesian closed - would you know of a reference for this? It would be good if there was some result that said categories internal to a suitable category E were cartesian closed.

Does anyone easily have a pointer?

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeNov 7th 2012

Does anyone easily have a pointer?

I don’t, but it doesn’t seem unreasonable to write out a proof for this type of thing. Without having written anything down, I’d guess that if $E$ is finitely complete and cartesian closed, then $Cat(E)$ is also finitely complete and cartesian closed. Then $n$-fold categories would be cartesian closed by induction.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeNov 7th 2012
• (edited Nov 7th 2012)

Well, I may as well have a go at writing down a sketch of a proof of the assertion from my previous comment, that if $E$ is finitely complete and cartesian closed, then $Cat(E)$ is also finitely complete and cartesian closed.

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

Now let $E$ be finitely complete and cartesian closed. Then $E^C$ is cartesian closed for any finite category $C$ (by adapting Mike’s proof, using only finite ends). This applies in particular to the case where $C$ is a suitable truncation of the simplex category, say where $C$ is opposite to the category of nonempty ordinals up to cardinality 3. Now $Cat(E)$ is a full subcategory of $E^C$, and it should be simple to see directly that it is an exponential ideal of $E^C$; in particular, it’s cartesian closed. (When I say “easy to see directly”, I mean that we don’t need to consider colimits in $Cat(E)$ or its being a reflective subcategory of $E^C$ – just use the formula for exponentials in $E^C$ to suggest the correct construction of exponentials in $Cat(E)$.)

Does this seem reasonable?

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeNov 7th 2012
• (edited Nov 7th 2012)

Maybe some of this could be copied to the entry n-fold category.

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeNov 7th 2012

The closest thing I can think of to a reference for this is the remark following B2.3.15 in the Elephant. But surely someone, somewhere, must have written it down before…

• CommentRowNumber6.
• CommentAuthorDavidRoberts
• CommentTimeNov 8th 2012

Ehresmann’s original proof, which I alluded to on the page internal category, is really quite awful. It’s working with generalised sketches, and for some reason either this precludes using the simple machinery from Todd’s #3, or Ehresmann just wasn’t in that frame of mind.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeNov 8th 2012

Can you add a pointer, or a page number, or something, to identify the alluded-to proof in Ehresmann’s work‘?

• CommentRowNumber8.
• CommentAuthorDavidRoberts
• CommentTimeNov 8th 2012

Can do.

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeNov 12th 2012

I got around to writing out a proof at internal category. Please feel free to check for accuracy.

If the ambient category $\mathcal{C}$ is a cartesian closed category, then the category $Cat(\mathcal{C})$ of categories internal to $\mathcal{C}$ is also cartesian closed.