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Given a category $\mathsf{C}$, there are two ways to define the notion of a group object in $\mathsf{C}$:

  1. As a quadruple $(G, m, i, e)$, where $G\in \mathsf{C}$ is some object, $m:G\times G\to G$ and $i:G\to G$ and $e:*\to G$ morphisms (where we are assuming both that $G\times G$ exists in $\mathsf{C}$ and that $\mathsf{C}$ admits a terminal object $*$), satisfying the usual axioms.
  2. As a lift of some representable functor $\mathsf{C}^\text{op}\to \mathsf{Set}$ to $\mathsf{C}^\text{op}\to \mathsf{Grp}$, i.e., as a functor $h:\mathsf{C}^\text{op}\to \mathsf{Grp}$ such that the postcomposition $U\circ h:\mathsf{C}^\text{op}\to \mathsf{Set}$ is representable, where $U:\mathsf{Grp}\to \mathsf{Set}$ is the forgetful functor.

For the first notion to make sense, we of course need $\mathsf{C}$ to admit (at least some) finite limits; conversely, when $\mathsf{C}$ is finitely complete (or even cartesian monoidal), then the two notions above are clearly equivalent (up to unique isomorphisms...) by Yoneda's Lemma. However, the second definition makes sense without any assumptions on $\mathsf{C}$ whatsoever.

My question is: can the second notion make sense without the first one doing so? Specifically, if $\mathsf{C}$ is a (locally small) category and $h:\mathsf{C}^\text{op}\to \mathsf{Grp}$ a functor such that the composite $U\circ h:\mathsf{C}^\text{op}\to \mathsf{Set}$ is representable, does this necessarily imply that $\mathsf{C}$ has a final object and a self-product of an object representing $U\circ h$?

I am having a very hard time coming with either a proof or a counterexample. The problem is that I do not know of any results that guarantee the existence of certain objects in a category, but it seems like condition (2) above places serious restrictions on the kind of category $\mathsf{C}$ is allowed to be. For instance, my first thought was to try the classifying category $\mathbb{B}G$ for a group $G$, but it is easy to show that the if $\mathsf{C}$ has only one object, then the above statement is true (the only such category that works is the "setoid" $\{*\}=\mathbb{B}\{e\}$). Besides, I am finding it challenging to conceptualize what could go on in such a situation; most natural categories I can think of which are not finitely complete do not seem to admit such functors. My hunch is that if even if there is some such category $\mathsf{C}$, then we can enlarge it appropriately to a minimal category (of say some presheaves on it) for which this bigger category admits the products which $\mathsf{C}$ does not see; but is there a general framework that such ideas fall into?

Thanks!

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  • $\begingroup$ You can pretty much define group objects in a monoidal category with diagonals, but the axiom that states $g^{-1}g=1$ is perhaps a little wonky. You would ask that the functions $G\to G$ given on generalised elements by $g^{-1}g$ and $gg^{-1}$ factor through the given $I\to G$. You need the diagonals to get these functions, but asking that $I$ is terminal gives you a cartesian monoidal category. You could ask that $I$ is sent to the terminal object of set, for instance. $\endgroup$ Commented Oct 20 at 2:54

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Let $C$ be a category in which $G$ is a group object of type (1). Let $S$ be a subset of $\text{Obj}(C)$ containing $G$, and let $C|_S$ be the full subcategory on $S$. It seems to me that the functor $C|_S \longrightarrow \text{Set}$ represented by $G$ still has a lift to $\text{Group}$ as in (2), but the product $G \times G$ need not be in $S$.

Am I missing something?

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    $\begingroup$ I think you are right; I was tripping myself up for no reason (although this is a somewhat tricky issue at first glance, I think). For instance, we can take $\mathsf{C}$ to be $\mathsf{Set}$, any sufficiently complicated group $G$, and $S:=\{G\}$. Then if $G$ is not something silly (e.g., I think having two distinct group automorphisms suffices), $G$ will not be its self product in the subcategory $\mathsf{C}|_S$ (slightly different that $G\times G$ not being in $S$ I think, but same vein). I think it is still worth having the question up; I imagine I am not the only person who runs into this. $\endgroup$ Commented Oct 20 at 3:10
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    $\begingroup$ As Dhruv’s comment points out, “there is no product $G \times G$ in $C |_S$” is in general a bit stronger than “$G \times G \notin S$”. Still, it’s not hard to find plenty of examples: e.g. take $G$ any ordinary finite group; then in this case, any product $G \times G$ in any full subcategory of sets must be the ordinary set product, just by checking the size of hom-sets. $\endgroup$ Commented Oct 20 at 8:49

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