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I would like to know the standard usage of "lax colimit" and "oplax colimit" in the 2-categorical literature. The nLab does not give an explicit definition of "lax colimit", as far as I can see, and I don't know what the most reliable source is. I think I have seen at least one paper using each convention, but I have not encountered the notion often enough to have a good sense of whether this one of those places where the terminology is not really standardized, or there is general agreement with a few exceptions.

Concretely, given a diagram X : I → C in a 2-category C (for my purposes indexed by a 1-category I), suppose I have a cone (Y, {gi}i∈Ob I, {αf}f∈Mor I), with gi : Xi → Y, and for f : i → j in I, αf : gjf → gi (such that various diagrams commute). Is this a lax colimit cone or a oplax colimit cone?

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This is probably making things worse, but personally I've never understood why (some) people say "oplax" rather than "colax". In a 2-category, "op" refers to reversing the 1-cells and "co" refers to reversing the 2-cells. Here, we're reversing a 2-cell. Hence, it should be "co". Moreover, in the 1-categorical world, the standard prefix for dualization is "co", not "op". –  Tom Leinster Mar 3 '10 at 0:54
I agree in principle, but carrying this logic to its conclusion, don't we find that "colimits" in a 1-category should be called "oplimits"? Alternatively, reverse the notation C^op and C^co throughout. I don't see either of these happening... –  Reid Barton Mar 3 '10 at 2:26
No, because reversing the (1-)morphisms in a (1-)category is called "co". So colimits are still colimits. Anyway, it's not that I'm a huge fan of the "op"/"co" system. It's that I don't understand why people choose to use the less common prefix ("op") when the more common one ("co") is logically correct. So, I merrily say "colax". –  Tom Leinster Mar 3 '10 at 3:36
If "in a 2-category 'op' refers to reversing the 1-cells" but "reversing the 1-morphisms in a 1-category is called 'co'," then wouldn't it follow that while we have limits and colimits in a 1-category, we would have limits and oplimits in a 2-category? But no one says that. It's true that in a 2-category, we write C^op to mean reversing the 1-cells and C^co for reversing the 2-cells, but in a 1-category we also write C^op for reversing the 1-cells, and nevertheless we use "co-" for most dualizations there. –  Mike Shulman Mar 3 '10 at 6:04
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2 Answers

up vote 5 down vote accepted

I agree with Finn that the way to derive the correct choice of lax vs oplax is to connect it back to natural transformations. Of course, as Finn pointed out, there is controversy over the choice for natural transformations, but my views on that are clear at the nlab page so I'll just write using that terminology.

However, in contrast to Finn, I think what you've got there is actually a lax cocone, since it is given by a lax natural transformation $* \to C(X-,Y)$, where $*:I^{op}\to Cat$ is the functor constant at the point. It's true that, as Finn says, it is also an oplax natural transformation from $X$ to $\Delta_Y$, where $\Delta_Y$ is the functor $I\to C$ constant at $Y$. But I think it's better to think of a cone as a transformation $* \to C(X-,Y)$, since this is the version that generalizes to weighted limits: for any weight $J:I^{op}\to Cat$, a $J$-weighted cocone is a transformation of the appropriate sort $J\to C(X-,Y)$.

The weighted-limit perspective on lax (co)limits is especially valuable because of the existence of lax morphism classifiers. Namely, for any weight $J$ there is another weight $J^\dagger$ such that lax transformations out of $J$ are the same as strict (or pseudo) transformations out of $J^\dagger$. Thus, lax $J$-weighted limits are the same as ordinary $J^\dagger$-weighted limits, so that lax $J$-weighted cones and limits are the same as ordinary $J^\dagger$-weighted cones and limits. Thus a "lax limit" is really just a particular type of weighted limit, whose weight happens to be of the form $J^\dagger$. Similarly, there is an oplax morphism classifier $J^\diamond$.

I think the choice I'm proposing is fairly widespread in Australia. For instance, it's the one used here and here and here. Actually, I'm not sure offhand whether I've even ever seen the other choice in print.

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Well, the smarmy answer is that it's neither, because you haven't given it a universal property. What you have is what I would call an oplax (co)cone, an 'oplax' transformation $X \Rightarrow \Delta_Y$. But there is no agreement in the literature over whether this sort of transformation (with the 2-cells in the naturality squares going 'upwards') should be called 'lax' or 'oplax'. See the discussion here, under '"Lax" versus "oplax"'.

Whichever you choose, the lax colimit of X will satisfy the 'usual' universal property, which I think is a natural equivalence $C(\operatorname{lcolim} X, A) \simeq \mathrm{Lax}(X, \Delta_A)$.

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Oops, I forgot that "lax colimit cone" means "cone which is actually a lax colimit". I just wanted to say "cone of the type appearing in the definition of lax colimit". –  Reid Barton Mar 3 '10 at 2:22
Connecting the (op)lax convention for colimits to the convention for natural transformations makes a lot of sense. But do you have a sense of how the terms are used in the literature? Do people who have the opposite convention for lax natural transformation generally also have the opposite convention for lax limits and colimits? –  Reid Barton Mar 3 '10 at 2:24
+1 Finn: that's a fantastically detailed n-Lab discussion that you linked to. I had no idea, in fact, that there was any controversy over the direction of lax transformations. Shows what I know. –  Tom Leinster Mar 3 '10 at 3:40
@Reid: I didn't really answer your question, did I? I think Mike's answer is better than mine, so I'd go along with him. In particular, I was probably wrong to suggest that (op)lax cones are defined simply by means of constant functors. For example, Johnstone, in the Elephant, uses the opposite convention for lax transformations, but he defines a lax cone as an oplax transformation out of a constant functor. He doesn't explain exactly why, but the ensuing Lemma 1.1.6 suggests that it's for the same reason that Mike gives. –  Finn Lawler Mar 3 '10 at 18:03
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