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According to https://ncatlab.org/nlab/show/delooping#delooping_of_a_group_to_a_groupoid we can think of delooping of a group as the one object groupoid $BG$ consisting of a single object and whose morphisms are the elements of the group $G$ and the composition of morphisms are given by the group multiplication.

Now let us consider a category $C$ with a group object $\bar{G}$.

Now if $C$ has a forgetful functor to $Sets$ then there is a natural way to associate a one object groupoid to the group object (For example the way we talk about delooping of a Topological group, Lie group..etc).

My question is the following:

Let $D$ be a category with a group object $G'$(without any other further assumptions). Is there an analogous way to associate a one object groupoid to $G'$ ?

Or is there any necessary and /or sufficient condition for doing so?

Apology in advance if my question is not making much sense.

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    $\begingroup$ You have to make some more requirements. Every object in an arbitrary category gives a one-object groupoid: its automorphism group. Since a one object groupoid is exactly the same info as a group, you might as well phrase your question as: given a category C, a group object G in C, how do I associate to G (presumably functorially) a plain group G' (i.e. a group object in Set)? $\endgroup$
    – David Roberts ♦
    Oct 18, 2020 at 7:43
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    $\begingroup$ If $C$ has pullbacks you can always see a group object as an 1-object groupoid object, by taking the groupoid object $G\rightrightarrows \ast$, and this is pretty much what's going on in the examples you're referring to. I'm not quite sure what else is there to say. $\endgroup$
    – Denis Nardin
    Oct 18, 2020 at 8:41
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    $\begingroup$ @DenisNardin you don't need pullbacks, just (enough) finite products, which you need to be able to define a group object. $\endgroup$
    – David Roberts ♦
    Oct 18, 2020 at 9:04
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    $\begingroup$ "I was asking that for a category C and a group object G in C can we always associate a BG like object(something like delooping) which will also be a one-object groupoid? " It will be a one-object groupoid internal to C. You must distinguish between a groupoid object in C, and a groupoid. It's not clear to me which one one you want, now, looking at the original question and your comments. "Is the groupoid object ... you mentioned in the previous comment....delooping ?" Yes. $\endgroup$
    – David Roberts ♦
    Oct 18, 2020 at 9:05
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    $\begingroup$ AdittyaChaudhuri: In order to say whether it is a delooping you need a notion of loop, which you haven't provided. But something like that is likely true (I think it's going to be a delooping in the nonabelian derived category of your ambient category, but I haven't thought this through). @DavidRoberts I just didn't want to spend time figuring out what a groupoid object is if you don't have enough pullbacks (I think the correct strategy is to work in $\mathcal{P}_\Sigma$, but I haven't thought it through) $\endgroup$
    – Denis Nardin
    Oct 18, 2020 at 9:12

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To provide a constructive answer, suppose $C$ is a category with finite products, and $G$ is a group object in $C$. Then the one-object groupoid object $G\rightrightarrows \ast$ in $C$ really is the delooping of $G$ in the 2-category $\mathbf{Gpd}(C)$ of groupoid objects1 in $C$. That is, there is a 2-commuting square making $G$ the "loop space object" of $G\rightrightarrows \ast$ at the canonical basepoint.

1 A groupoid object in a category without all finite limits is the same as in a finitely-complete category, just that one must demand that the relevant pullbacks exist, or alternatively, one supplies explicit objects of composable pairs and triples. If this makes you uncomfortable, assume $C$ has all finite limits instead. Personally, I would go so far as to only require a terminal object exists, in order to define group objects.

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  • $\begingroup$ Thank you very much for the answer. Yes I was saying about the one-object groupoid internal to $C$ in the above comments. I am apologising for the confusion. $\endgroup$
    – Adittya Chaudhuri
    Oct 18, 2020 at 10:02

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