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We want to show that if is a bicategory then is the free 2-cocompletion of .
There are several kinds of 2-colimit that we’ll need to talk about. Let and be pseudofunctors. Then
The 2-colimit satisfies
This is what in the literature is often called a bilimit.
If is a strict 2-category, the pseudocolimit satisfies the same property up to isomorphism.
If is strict and and are strict 2-functors, then the strict pseudocolimit satisfies
where on the right the functor category is that of strict 2-functors, pseudonatural transformations and modifications.
Under the same hypotheses, the strict colimit satisfies
where now denotes the category of strict 2-functors and strict transformations (and modifications).
We need to show that has all small 2-colimits:
is strictly 2-cocomplete: its underlying 1-category has small colimits, and is enriched and tensored over itself, so that it has strict -weighted colimits.
Pseudocolimits, strict or otherwise, are a fortiori 2-colimits, and strict pseudocolimits are just strict colimits whose weights are ‘cofibrant’ in a suitable sense. Moreover, if is a strict 2-category, then for any index bicategory there is a strict 2-category such that strict functors are the same thing as pseudofunctors , and the 2-colimit of pseudofunctors is equivalent to the strict pseudocolimit of the strictified functors. So a strictly 2-cocomplete strict 2-category is also 2-cocomplete.
is therefore a strict 2-category, and it has all non-strict strict 2-colimits. ( We can now try to compute colimits pointwise in -weighted) colimits, which are computed pointwise as usual for enriched strictly-enriched functor categories: if now and are strict, then we set can define , and verify its universal property: property follows:
So has 2-colimits.
So Finally, we need to show that if has is strict a 2-colimits cocomplete and bicategory, hence then also there non-strict is 2-colimits. a 2-equivalence
Finally, we need to show that if is a cocomplete bicategory, then there is a 2-equivalence
For this we simply follow the usual reasoning: from left to right we compose with the Yoneda embedding , and given a functor we get a cocontinuous sending to .
For The this co-Yoneda we lemma simply shows follow that the every usual reasoning: from left to right we compose with the Yoneda embedding , and given if a functor we is get a cocontinuous then , sending showing that the functor to is essentially surjective. It is 2-fully-faithful by the universal property of colimits: a transformation gives rise to an essentially unique transformation .
The co-Yoneda lemma shows that every , and if is cocontinuous then , showing that the functor is essentially surjective. It is 2-fully-faithful by the universal property of colimits: a transformation gives rise to an essentially unique transformation .