3
$\begingroup$

There seems to be many ways to obtain a 1-category out of a 2-category:

  1. 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.
  2. 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.
  3. 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.

$\endgroup$
3
  • $\begingroup$ How is (3) different from (2)? $\endgroup$
    – Qfwfq
    Dec 25, 2017 at 21:54
  • $\begingroup$ @Qfwfq it depends with respect to which interval you compute π_0 $\endgroup$ Dec 26, 2017 at 10:20
  • $\begingroup$ @Qfwfq in principle, you are quotienting for two different equivalence relations. $\endgroup$
    – fosco
    Dec 26, 2017 at 18:51

1 Answer 1

2
$\begingroup$
  1. 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.)

  2. 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.

  3. 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 $K^c$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.