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In what sense does a functor "lift/create/detect" limits?

 
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Last Post by Sayantan Roy 5 years ago
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 Sayantan Roy
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Topic starter 05/12/2020 7:54 am   [#34]

This question has its motivations in trying to understand more deeply the guiding intuitions for coining these terms. I am using the definitions of Joy of Cats.

A functor $F:\mathbf{A}\to\mathbf{B}$ is said to,

  • lift limits provided that for every diagram $D : \mathbf{I}\to \mathbf{A}$ and every limit $\mathcal{L}$ of $F \circ D$ there exists a limit $\mathcal{L}'$ of $D$ with $\mathscr{F}(\mathcal{L}') = \mathcal{L}$.
  • create limits provided that for every diagram $D : \mathbf{I}\to \mathbf{A}$ and every limit $\mathcal{L}$ of $F \circ D$ there exists a unique source $\mathcal{S}$ in $\mathbf{A}$ with $F(\mathcal{S})=\mathcal{L}$ and that, moreover, $\mathcal{S}$ is a limit $D$.
  • detect limits provided that a diagram $D : \mathbf{I}\to \mathbf{A}$ has a limit whenever $F\circ D$ has one.

I would like to know for example,

  1. In what sense does $F$ "lift" limits? 
  2. In what sense does $F$ "create" limits? 
  3. In what sense does $F$ "detect" limits?

It may look silly to try to understand terms in this way. An alternative is just to grasp the concept itself and forget about the terminology. However, many times I have seen that trying to understand the terminology often rewards back with beautiful and deep intuitions. I hope it turns out like that in this case as well. Apologies if that doesn't turn out to be so. 

 

 

 

 

 



   
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 Joshua Wrigley
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06/12/2020 2:18 pm  

Hi, the wording of "in what sense" is a bit vague so I'm going to interpret your question as why have the English words "lift", "create" and "detect" been used.  The terminology derives from the case where we have a category {\bf B} that we understand well, and a category {\bf A} which we don't.  If we have a suitably well behaved functor F \colon {\bf A} \to {\bf B} we can hope to understand the behaviour of limits of {\bf A} in terms of the limits of {\bf B}. Firstly, clearly 2) \implies 1) \implies 3).

3) We say that F detects limits because we can detect if a limit exists in {\bf A} by considering the image in {\bf B}.

1) We say that F lifts limits because the limit structure in the image of F lifts to structure on {\bf A}, so we can reason about {\bf A} in {\bf B}.

2) We say that F creates limits because the data of F uniquely determines limits in {\bf A}.

The classic example is given by forgetful functors U \colon {\bf A} \to {\bf Sets} into the category of sets.  Consider the categories taken from [SGL, pg.62] (both topoi) {\bf B}G and {\bf B}G^{\delta}, that is the continuous and discrete representations of a topological group G.  Then the forgetful functors U^1 \colon {\bf B}G \to {\bf Sets} and U^2 \colon {\bf B}G^\delta \to {\bf Sets} both vacuously detect limits (as the categories {\bf Sets}, {\bf B}G and {\bf B}G^{\delta} have all limits).  Also, U^2 creates limits but U^1 does so if and only if the intersection of any collection of open subgroups is again open.



   
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 Sayantan Roy
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Topic starter 06/12/2020 4:48 pm  

I think this is exactly what I was looking for. However, although (3) and (1) makes sense to me, I am not sure (2) does justice to the term "create" because $F$ doesn't really create anything new in $\mathbf{A}$. But anyway, it's a very nice explanation. Many thanks. 



   
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 Joshua Wrigley
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06/12/2020 9:05 pm  

I guess whether F "creates" limits depends on whether those limits exist from before or after you observe them.



   
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 Sayantan Roy
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Topic starter 09/12/2020 6:19 am  

Probably, but I still can't shake the feeling of a lack of natural terminology to express this concept (if I am not missing anything). An idea (motivated by a private chat with Aleks Kissinger) is to consider "functorial equations". More specifically, if $F:\mathbf{A}\to\mathbf{B}$ is a functor then for $F$ all these concepts essentially reduces to finding certain "nice" solutions of the "functorial equation" of the form $F(\mathcal{X})=\mathcal{Y}$ (Aleks Kissinger told me to consider isomorphism instead of equality, but equality suffices for this case) for "nice" $\mathcal{Y}$.  Here $\mathcal{X},\mathcal{Y}$ are respectively $\mathbf{A}$-source and $\mathbf{B}$-source variables (admittedly, all these are pretty vague right now, but I hope that I am able to convey the big picture).

I believe this goes well to what you have proposed in the beginning. Here $\mathcal{X}$ is the "unknown" and $\mathcal{Y}$ is the "known". 

 

 

 



   
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