2
$\begingroup$

A quick question about lax co/limits.

Strictly, when $F : J\to \bf A$ is a diagram and $J$ has an initial object $\varnothing$, then $\varprojlim F \cong F(\varnothing)$; dually, if $\cal J$ has a terminal object, then $\varinjlim F\cong F(*)$.

If $F$ is a diagram between 2-categories (same notation), and $J$ has a lax initial, lax terminal object, is a similar statement true for lax co/limits?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

No. If $\mathcal{J}$ is the interval category (two objects and one nonidentity morphism between them), which certainly has an initial and a terminal object, then the lax limits and colimits of such a diagram are comma objects of the morphism that determines its image, neither of which is usually equivalent to its domain or codomain.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I agreed the first time I read this, but then I thought: isn't the comma object the lax limit of a cospan? $\endgroup$
    – fosco
    Commented Feb 20, 2017 at 11:05
  • $\begingroup$ No, the lax limit of a cospan is different from the comma object; it involves two nonidentity 2-cells, whereas the comma object involves only one. $\endgroup$
    – Mike Shulman
    Commented Feb 20, 2017 at 11:53
  • $\begingroup$ Ah, the nLab page solved the terminology issue that you raised (or that I misinterpreted until today). The lax co/limit of $f : X \to Y$, seen as a diagram $\Delta[1] \to\cal A$, is the comma of $f$ and the identity (which is the way I interpreted your answer at first sight). So, thanks again! $\endgroup$
    – fosco
    Commented Feb 20, 2017 at 21:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .