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It's not hard to see that a category is finitely complete if it has finite products and equalizers. In short, this is because one can write all limits as iterations of these two "operations".

I wonder if there is a 2-version of this. In particular,

Does a category have all finite 2-limits if it has all 2-equalizers and 2-products?

My instinct is no, and that we will need another(or several more) limits to build all 2-limits.

Of course the question can be generalized to n-limits, and I'm curious about that also.

edited Oct 12 2011 at 4:07

Finn Lawler
2,361●1●5●11

asked Oct 12 2011 at 2:14

B. Bischof
2,621●22●34

The nlab has an overview over some classes of 2-limits which do not arise for 1-categories, for example inserters. You cannot generate them from 2-equalizers or 2-products. – Martin Brandenburg Oct 12 2011 at 12:23

Finn Lawler · Answer 1 · 2011-10-12 04:05:51Z UTC

Your suspicion is correct: in general, a V-category has all weighted V-limits if it has all conical V-limits and is cotensored over V (see Kelly's Basic Concepts of Enriched Category Theory, section 3.10). For V = Cat (and this is true for bicategories too), cotensors can be constructed from conical limits and cotensors with the arrow category 2, although I don't know the original reference for that. 'Finite' limits are a bit more complicated in the enriched case, but see Street's 'Limits indexed by category-valued 2-functors', JPAA 1972, for the 2-case.

2-completeness analog of completeness theorem

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