Category Theory
Zulip Server
Archive

You're reading the public-facing archive of the Category Theory Zulip server.
To join the server you need an invite. Anybody can get an invite by contacting Matteo Capucci at name dot surname at gmail dot com.
For all things related to this archive refer to the same person.

Stream: theory: category theory

Topic: grothendieck construction as a colimit

Matteo Capucci (he/him) (Apr 19 2021 at 07:02):

On the nLab page for the Grothendieck construction, it's stated $\int F$ is an oplax colimit. The way it is phrased makes me think it refers to the covariant Grothendieck construction. Is there a similar characterization for the contravariant Grothendieck construction? Is it just the lax colimit of a contravariant pseudofunctor $F$?

Amar Hadzihasanovic (Apr 19 2021 at 07:34):

You get the contravariant one from the covariant one by applying $-^\mathrm{op}$ on $\mathbf{Cat}$.

You start from $F: C \to \mathbf{Cat}$. First you obtain the covariant construction $\int F$ as an oplax colimit, i.e. a universal lax cone under $F$ with tip $\int F$. Applying the involution $-^\mathrm{op}$ sends this to a universal oplax cone under $F^\mathrm{op}$ (the “pointwise opposite” of $F$) with tip $(\int F)^\mathrm{op}$, which is the contravariant Grothendieck construction. So the latter should be the lax colimit of the pointwise opposite of $F$.

Matteo Capucci (he/him) (Apr 19 2021 at 08:36):

Right, I see :thinking:

Matteo Capucci (he/him) (Apr 19 2021 at 08:39):

So if I want to get $\int_{op} F$ (I use $\int_{op}$ to denote contravariant Grothendieck), I just do $\int F^{pop}$ (pop means pointwise opposite), since $(F^{pop})^{pop} = F$

Amar Hadzihasanovic (Apr 19 2021 at 08:59):

No, that's not right. It's $\int_\mathrm{op} F = (\int F)^\mathrm{op}$.

Amar Hadzihasanovic (Apr 19 2021 at 08:59):

Which is different from $\int (F^\mathrm{pop})$.

Amar Hadzihasanovic (Apr 19 2021 at 09:01):

You can see that the latter is the oplax colimit of $F^\mathrm{pop}$, while earlier I characterised $\int_\mathrm{op} F$ as its lax colimit.

Matteo Capucci (he/him) (Apr 19 2021 at 09:40):

Amar Hadzihasanovic said:

No, that's not right. It's $\int_\mathrm{op} F = (\int F)^\mathrm{op}$.

We might be working with different definitions then.

Matteo Capucci (he/him) (Apr 19 2021 at 09:41):

Let me check

Matteo Capucci (he/him) (Apr 19 2021 at 09:42):

Mmh so here $\int_{op} F^{pop} = (\int F)^{op}$

Matteo Capucci (he/him) (Apr 19 2021 at 09:43):

But $F : C^{op} \to Cat$

Amar Hadzihasanovic (Apr 19 2021 at 10:38):

The variance of $F$ is irrelevant (you can always just let $D = C^\mathrm{op}$). What matters is that the covariant construction for $F: C \to \mathbf{Cat}$ produces an opfibration over $C$ and the contravariant construction a fibration over $C^\mathrm{op}$. If we fix what the covariant one is and call it $\int F$, then I guess that $(\int F)^\mathrm{op}$ and $(\int F^\mathrm{pop})^\mathrm{op}$ are equally good.

Amar Hadzihasanovic (Apr 19 2021 at 10:39):

If you adopt the latter, then I guess it does make the “colimit” characterisation simpler: $(\int F^\mathrm{pop})^\mathrm{op}$ is the lax colimit of $F$.

Amar Hadzihasanovic (Apr 19 2021 at 10:40):

This is all assuming that you're right that (the covariant construction) $\int F$ is the oplax colimit of $F$, which I have not checked :D

Matteo Capucci (he/him) (Apr 19 2021 at 15:06):

The nLab says so :) but it seems very reasonable up to what oplax/lax corresponds to

Matteo Capucci (he/him) (Apr 19 2021 at 15:06):

Amar Hadzihasanovic said:

If we fix what the covariant one is and call it $\int F$, then I guess that $(\int F)^\mathrm{op}$ and $(\int F^\mathrm{pop})^\mathrm{op}$ are equally good.

I don't understand this

Matteo Capucci (he/him) (Apr 19 2021 at 15:07):

What do you mean by equally good?

Matteo Capucci (he/him) (Apr 19 2021 at 15:21):

I think I'm challenging your claim that $(\int F)^{op} = \int_{op} F$. In the first category, morphisms $(A,A') \to (B,B')$ are given by arrows $f : B \to A$ and $f' : F(f)(B') \to A'$. In the second category, morphisms $(A,A') \to (B,B')$ are given by arrows $f : B \to A$ and $f' : A' \to F(f)(B')$.
If you flip $F$ pointwise, then the latter arrow lives in $F(A)^{op}$, and thereby corresponds to an arrow $F(f)(B') \to A'$ in $F(A)$.

Matteo Capucci (he/him) (Apr 19 2021 at 15:27):

Your suggestion is valuable anyway: $op$ exchanges lax and oplax co/limits, so if $\int F$ is the oplax colimit of $F$ then $(\int F)^{op}$ has to be the lax colimit. But then $(\int F)^{op} = \int_{op} F^{pop}$. So the lax colimit of $F$ is presented by the contravariant Grothendieck construction of its pointwise opposite.

Matteo Capucci (he/him) (Apr 19 2021 at 15:30):

Now $\int_{op} F = (\int F^{pop})^{op}$. The latter is the lax colimit of $F^{pop}$, hence $\int_{op} F$ presents the lax colimit of its pointwise opposite.

Matteo Capucci (he/him) (Apr 19 2021 at 15:33):

What's the relationship between op/lax co/limits of $F$ and those of its ptwise opposite then? :thinking:

Amar Hadzihasanovic (Apr 19 2021 at 18:37):

@_Matteo Capucci (he/him)|275932 said:

What do you mean by equally good?

I mean that the question “which one is the (contravariant) Grothendieck construction” seems equally unimportant to me as “is $i$ or $-i$ the root of -1”. The contravariant Grothendieck construction is, in its essence, a way of turning a pseudofunctor $C \to \mathbf{Cat}$ into a Grothendieck fibration over $C^\mathrm{op}$. Since there is an involutive automorphism on pseudofunctors $C \to \mathbf{Cat}$, the “pointwise opposite”, this way “is” two ways, related by the automorphism, and privileging one is only a matter of convenience. I have found $(\int F)^\mathrm{op}$ more useful in my own work so I tend to think of it as “the” one but that's completely arbitrary.

Amar Hadzihasanovic (Apr 19 2021 at 18:44):

There's a similar argument to be made about whether the covariant or the contravariant Grothendieck construction are “the Grothendieck construction”... really it's all one thing, with a $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ group of automorphisms :)

Reid Barton (Apr 19 2021 at 19:24):

It seems sensible to set things up to be "unital" in the sense that if $C = 1$, then both Grothendieck constructions applied to $F : 1 \to \mathrm{Cat}$ produce the original category $F(*)$, not its opposite; particularly if one wants the construction to be given by a lax (or oplax) colimit.

Reid Barton (Apr 19 2021 at 19:25):

If I'm not mistaken, only one of the choices being discussed here has that property.