Can "equipping" a set with structure be given a precise definition?

Question

An algebraic structure such a group can be formally defined as an ordered pair $(G,\cdot)$, where $G$ is a set and $\cdot$ is a binary operation on $G$ satisfying various axioms. This is not the only way to define groups: for instance, a group may be defined as an ordered quadruple $(G,\cdot,^{-1},e)$, where $\cdot$ is a binary operation (on $G$), $^{-1}$ is a unary operation, and $e$ is a constant; this definition is more in line with universal algebra.

However, my experience is that in practice, algebraists tend not to use either definition in their mental picture of groups. Instead, a group is described more informally as a set "equipped", "endowed", or "enriched" with some extra structure. The same goes for other algebraic structures of any kind. This way of thinking is reflected by how, in practice, algebraic structures are almost always referred to by the name of their underlying set.

Usually, the formal way to make sense of this "equipping" is just to define algebraic structures as $n$-tuples with certain properties. But I wonder if there is a way of defining them which feels closer to how they are actually conceived of. In particular, I would like to be able to speak of "elements of a group/ring/vector space" directly. In the usual formalism, that phrase has to be interpreted as a shorthand for "elements of the underlying set of a group/ring/vector space"; can this be avoided?

Answer 1

As an exercise in precisely precise language this is an interesting question (though not in the context of actually communicating mathematics with other people, where "elements of an object" is completely understood even if formally the object is a tuple rather than the set).

One way of resolving it would be to define a "group" (for example) as the set $G$ while defining a "group structure" as the quadruple $(G,\cdot,{}^{-1},e)$. "Equipping a set with a group structure" would then mean finding a group-structure quadruple whose first element is the set.

Answer 2

Possibly "equipping" a set $G$ with additional structure could be construed categorically as a map/functor/whatever from the category with one object, the set $G$, and only the identity self-map morphism, to whatever ("richer") category we want?

(This does not seem to require that $G$ be a set, though the ramifications of this non-requirement are completely unclear to me at the moment.)

Answer 3

You have a forgetful functor from $Grp$ (or $Ring$ or whatever) to $Set$, and a free functor adjoint to it.

This free functor from $Set$ to $Grp$ will be monadic and will admit an algebra, which will recover your original group. So up to the embedding of the free functor, the algebra is what equips your set with structure. (But by quotienting a free structure instead of adding stuff.)

Edit: A nice thing about this is that it generalizes to other forgetful functors. e.g: you may consider adding structure to a monoid or group so it becomes a ring, etc.

Answer 4

In category theory, groups and sets are organized in categories $\mathrm{Grp}$ and $\mathrm{Set}$, respectively. A set $X$ can be identified with a functor $X:1\to\mathrm{Set}$ and a group with a functor $G:1\to\mathrm{Grp}$, where $1$ is the terminal category. There is a functor $U:\mathrm{Grp}\to\mathrm{Set}$ which sends any group to its underlying set. In this context, equipping a set $X$ with a group structure is the same as finding a lift of $X:1\to\mathrm{Set}$ to $\mathrm{Grp}$ along $U$. I.e. finding a functor $G:1\to\mathrm{Grp}$ such that $UG=X$. Usually we allow for isomorphisms of functors instead of equality, so we can replace the last condition with $UG\simeq X$.