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I was wondering if we can construct the category $\mathbf{Grp}$ of groups from the function $G$ which associates to every set $X$ the set $G(X)$ of group structures on $X$, or what we have to add to this function to make it work.

Namely, a group is just a pair $(X,a)$ where $X$ is a set and $a \in G(X)$. The homomorphisms make some trouble, though. A homomorphism $(X,a) \to (Y,b)$ should be a map $f : X \to Y$ such that some property is satisfied, but I don't see how to express it in terms of $G(X)$ and $G(Y)$ alone. If $f$ is bijective, then we have an induced map $f_* : G(X) \to G(Y)$, and the condition is just $f_*(a)=b$. For general $f$, it seems that we need some extra structure on $G(-)$ to make it work. So my question is: What is this extra structure?

Of course, $G(X)$ is just the fiber of the forgetful functor $U : \mathbf{Grp} \to \mathbf{Set}$ at $X$, but I wonder if we can define this structure without "assuming" $\mathbf{Grp}$ is already there. The problem is that $U$ is an iso-fibration, but not a fibration or an opfibration. Let me know if I should clarify the question further.

The analogous problem for $\mathbf{Top}$ can be solved as follows: Consider the functor $T : \mathbf{Set}^{\mathrm{op}} \to \mathbf{Pos}$ which maps a set $X$ to the partially ordered set $T(X)$ of topologies on $X$ and a map $f : X \to Y$ to the map $f^* : T(Y) \to T(X)$, $f^*(\tau) = \{f^{-1}(U) : U \in \tau\}$. Then, a topological space is a pair $(X,\tau)$ with $\tau \in T(X)$, and a continuous map $(X,\tau) \to (Y,\sigma)$ is a map $f : X \to Y$ such that $f^*(\sigma) \subseteq \tau$. In other words, $\mathbf{Top}$ is the Grothendieck construction of the functor $\mathbf{Set}^{\mathrm{op}} \to \mathbf{Pos} \hookrightarrow \mathbf{Cat}$.

Martin Brandenburg
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    Just a (hopefully not so trivial) remark: if we instead wanted to do this for sets of semigroup structures $S(X) \subset X^{X \times X}$, then defining $S(f)(a) = a \circ f \times f$ we would get a functor $S \colon \mathsf{Set}^{op} \to \mathsf{Set}$ and I think that $\mathsf{Semigrp}$ is the category of elements of $S$, no? – qualcuno Jul 24 '21 at 17:03
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    @guidoar This only works for $A^{X \times X}$ for a fixed set $A$. – Martin Brandenburg Jul 24 '21 at 19:57
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    Ah, right, my bad. I suppose one could restrict to isos and do something like $a \mapsto f^{-1} \circ a \circ f \times f$, but this is already covered in the original question. – qualcuno Jul 24 '21 at 20:16
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    If you think of the group structure as the "graph" of the multiplication function (so, a subset of $G\times G\times G$, where $(a,b,c)$ lies in the subset if and only if $ab=c$ in $G$), then couldn't you define morphisms $f\colon (G,\mu)\to (K,\nu)$ as functions $f\colon G\to K$ with $f(\mu)\subseteq \nu$ (so that $(a,b,c)\in \mu$ must map to $(f(a),f(b),f(c))\in\nu$, which means that if $ab=c$ in $G$, then $f(a)f(b)=f(c)$ in $K$)? You would be using the fact that a function between group is a group homomorphism if and only if it is a semigroup homomorphism... – Arturo Magidin Jul 24 '21 at 23:48
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    @ArturoMagidin Oh, this looks good. Can you post this as an answer? The only thing which is missing is that $f(\mu)$ is not necessarily a group structure. – Martin Brandenburg Jul 25 '21 at 00:40
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    @MartinBrandenburg: Well, it's not necessarily a group structure on $K$, but it would be a group structure on a subset of $K$ (namely, $f(G)$). Don't know if that can be shoehorned into what you want, though. – Arturo Magidin Jul 25 '21 at 01:02
  • Wouldn't this fit on MO? – YCor Jul 27 '21 at 17:04
  • Here is a silly observation: The forgetful functor from the Krause--Nikolaus model category of groups (rather: group presentations) to sets is a fibration, so something must be lost on the way passing to the homotopy category (of classical groups). So perhaps the extra structure that's missing is some sort of generating set / presentation? – Ben Oct 12 '23 at 06:55

1 Answers1

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I think the following works.

Notice that a map between the underlying sets of two groups is a homomorphism iff its graph is a subgroup of the product $(\ast)$. We can make use of this.

The extra structure on $G(X) := \{\text{group structures on } X\}$ is the following:

There is a natural map $1 \to G(1)$, since the singleton has a canonical group structure, and natural maps $$G(X) \times G(Y) \to G(X \times Y),\quad (a,b) \mapsto a \times b,$$ since we can build the product of two groups. Next, for every injective map $i : X \to Y$ we have a partial map $$i^* : G(Y) \dashrightarrow G(X).$$ It is defined on the set of group structures on $Y$ for which $i$ (read this as $i(X) \subseteq Y$ if you like) is a subgroup, and simply restricts the group structure. If $i$ is bijective, then $i^*$ is total, and $\mathrm{id}^* = \mathrm{id}$.

But this is not a plain functor. If $i : X \to Y$, $j : Y \to Z$ are two injective maps, then we only have $i^* \circ j^* \leq (j \circ i)^*$. Here, we define $f \leq g$ for two parallel partial maps if $g(x)$ is defined as soon as $f(x)$ is defined and then $f(x) = g(x)$.

Abstractly, consider the category $\mathbf{Set}_i$ of sets with injective maps. It becomes a monoidal category with the cartesian product. Also, consider the category $\mathbf{Par}$ of sets and partial maps. It also becomes a monoidal category with the cartesian product (notice that this is not the categorical product here, which is $(X \times Y) \sqcup X \sqcup Y$, which I find very curious). Under the usual equivalence to $\mathbf{Set}_*$, this is the smash product. Actually, $\mathbf{Par}$ is a $\mathbf{Pos}$-enriched category with the definition $\leq$ above, hence a $2$-category.

The extra structure defined above says that $G$ is a lax monoidal oplax $2$-functor $\mathbf{Set}_i^{\mathrm{op}} \to \mathbf{Par}$.

Now, if $(X,a)$ and $(Y,b)$ are two groups, we define a homomorphism $f : (X,a) \to (Y,b)$ to be a map $f : X \to Y$ with the property that $a \times b \in G(X \times Y)$ belongs to the domain of definition of the partial map $(X,f)^* : G(X \times Y) \to G(X)$.

This is really just saying $(\ast)$ in a more abstract way. To verify that this defines a category, we need more axioms for $G$, though. For example, for the identity we need that $\Delta^* : G(X \times X) \to G(X)$ is defined for all $a \times a$.

Martin Brandenburg
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