Co/completeness of truncated 2-category There seems to be many ways to obtain a 1-category out of a 2-category:


*

*Dumb truncation. $\delta: 2\text{-Cat} \to \text{Cat}$ sends a 2-category $\cal K$ into the 1-category obtained forgetting the 2-cells. Only works with strict 2-categories.

*Core truncation. $c : 2\text{-Cat} \to \text{Cat}$ sends a 2-category $\cal K$ into the 1-category obtained taking isomorphism classes of 1-cells. Apparently this works also with bicategories.

*Geometric truncation. $\pi_{0*} : 2\text{-Cat} \to \text{Cat}$ sends a 2-category $\cal K$ into the 1-category obtained applying hom-wise the $\pi_0$ functor (so takes connected components of ${\cal K}(x,y)$).


Under which conditions is ${\cal K}^\delta, {\cal K}^c, {\pi_{0*}\cal K}$ a co/complete 1-category?
Non-strictness is not a big deal, I'm fine with strict 2-categories.
 A: *

*Every strict conical 2-limit in $K$ is also a 1-limit in $K^\delta$.  So if $K$ has all of those, then $K^\delta$ is complete.  (This is a special case of a general fact about enriched categories, since $K^\delta$ is the underlying ordinary category of the $\mathrm{Cat}$-enriched category $K$ in the sense of enriched category theory.)

*This is a 2-dimensional version of taking homotopy categories, which rarely turn out to be complete or cocomplete.  Some 2-limits, like 2-products, in $K$ become 1-limits in $K^c$.  Others, like pseudopullbacks, become weak 1-limits (satisfying existence but not uniqueness of factorizations) in $K^c$.  But even if $K$ is complete and cocomplete, like $\mathrm{Cat}$, $K^c$ need not have strict 1-pullbacks or 1-equalizers; see for instance here.

*This is basically a generalization of (2), since if $K$ is locally groupoidal then $K^c = \pi_{0\ast}K$ and in general $K^c$ is $\pi_{0\ast}$ of the local core of $K$, while $\pi_{0\ast}K$ is $(K')^c$ where $K'$ is $K$ with all 2-cells formally inverted.  So 2-products in $K$ become 1-products in $\pi_{0\ast}K$, while in general $\pi_{0\ast}K $ will not be complete or cocomplete even if $K$ is.  I don't offhand know whether there is anything analogous for $\pi_{0\ast}K$ to the weak pullbacks in $K^c$.
