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Stream: learning: questions

Topic: ✔ (co)limits in the covariant Grothendieck construction

Naso (May 13 2023 at 06:11):

nLab has a section on existence and characterization of (co)limits in a Grothendieck construction.

I believe this refers to the contravariant GC. What is the corresponding theorem for the covariant GC?
There are two references mentioned there which I looked at but couldn't find the answer in them...

Dylan Braithwaite (May 13 2023 at 10:41):

You can relate the contra- and covariant grothendieck constructions by $\int_\text{co}^{C \in \mathcal C} F(C) = \left(\int_\text{contra}^{C \in \mathcal C^\text{op}} F(C)^\text{op}\right)^\text{op}$. So using the fact that colimits in $\mathcal C$ are limits in $\mathcal C^\text{op}$. I think you should be able to write down a corresponding version of the existence theorems

Dylan Braithwaite (May 13 2023 at 10:49):

For example, this tells us $\int_\text{co}F$ is complete iff $\int_\text{contra}F^\text{ptwise-op}$ is cocomplete, so you get a sufficient condition for $\int_\text{co}F$ in terms of the cocompleteness of $\mathcal C^\text{op}$ and $F^\text{ptwise-op}$.

Naso (May 13 2023 at 12:40):

Dylan Braithwaite said:

For example, this tells us $\int_\text{co}F$ is complete iff $\int_\text{contra}F^\text{ptwise-op}$ is cocomplete, so you get a sufficient condition for $\int_\text{co}F$ in terms of the cocompleteness of $\mathcal C^\text{op}$ and $F^\text{ptwise-op}$.

Thank you, Dylan. But I'm still getting confused by all these op's. I am thinking about applying the covariant GC to the same contravariant functor, $F : \mathcal{C}^{op} \to \mathcal{Cat}$,. I'm not sure if your description assumes $F$ has domain $\mathcal{C}$ or $\mathcal{C}^{op}$.

So given $F : \mathcal{C}^{op} \to \mathcal{Cat}$, applying the covariant GC I get $\int_\text{co}^{C \in \mathcal{C}^{op}} F$ is complete if

1. $\mathcal{C}$ is cocomplete
2. $F(J)$ is cocomplete
3. $F(f) : F(J) \to F(K)$ preserves colimits

..? I'm really not sure.

Dylan Braithwaite (May 13 2023 at 14:28):

I was assuming $F$ with domain $\cal C$, since thats the usual data for the covariant GC, but lets step through it with the domain as $\cal C^\text{op}$.

So for $F : \cal C^\text{op} \to \sf Cat$ and $F^\mathrm{p}$ the pointwise opposite of $F$, we have that the following are equivalent:

• the covariant Grothendieck construction $\int^{\cal C^\text{op}}_\text{co} F$ is complete
• the opposite of the contravariant Grothendieck construction $\left(\int^{\cal C}_\text{contr} F^\mathrm{p}\right)^\text{op}$ is complete
• the contravariant Grothendieck construction $\int^{\cal C}_\text{contr} F^\mathrm{p}$ is cocomplete

From the result on the nlab, the Grothendieck construction $\int^{\cal C}_\text{contr} F^\mathrm{p}$ is cocomplete if:

• $\cal C$ is cocomplete
• $F^\mathrm{p}(C)$ is cocomplete for all $C \in \cal C$
• $F^\mathrm{p}(f) : F^\mathrm{p}(C’) \to F^\mathrm{p}(C)$ has a left adjoint for all $f \in \mathcal{C}^(C, C’)$

Expanding the definition of $F^\text{p}$, this is equivalent to asking that:

• $\cal C$ is cocomplete
• $F(C)^\text{op}$ is cocomplete for all $C \in \cal C$
• $F(f)^\text{op} : F(C’)^\text{op} \to F(C)^\text{op}$ has a left adjoint for all $f \in \mathcal{C}(C, C’)$

and one more step of simplification:

• $\cal C$ is cocomplete
• $F(C)$ is complete for all $C \in \cal C$
• $F(f) : F(C’) \to F(C)$ has a right adjoint for all $f \in \mathcal{C}(C, C’)$

Naso (May 13 2023 at 23:01):

Thank you! :blush: The one example I was looking at is in posets where completeness and cocompleteness is the same, so that didn't help me work this one out besides the 3rd point (has a right adjoint). Interestingly, it did show me that this condition is not necessary, since our (covariant) GC poset is cocomplete but the maps $F(f)$ do not preserve limits/have a left adjoint.

Notification Bot (May 13 2023 at 23:01):

Naso has marked this topic as resolved.