# Soft question on universal properties

My question heads towards an intuition of universal properties. Take for example a group $$G$$ and a subset $$M$$. Then $$F_M$$, the free group on $$M$$ together with a map of sets $$i:M\to U(F_M)$$ where $$U$$ denotes the forgetful functor, is defined by the universal property that any map of sets $$M\to U(H)$$ factors through $$U(F_M)$$ for a group $$H$$. A concrete model for the free group is given by the free abelian group with basis $$M$$. Although I understand that universal properties fix objects up to isomorphism I am not sure about the following two things:

1) The explicit construction of $$F_M$$ clearly reveals it to be the smallest subgroup of $$G$$ containing $$M$$ in the sense that it is the group generated by $$M$$. How could this fact be concluded from the universal property?

2) More generally: Universal objects are often charaterized as being the smallest or most general objects regarding a certain property. I understand this in examples like the free group, the completion of a topological space, the tensor algebra, etc. However, I wonder if being the smallest or most general can be expressed (or even defined) in a precise categorical sense?

Concerning the last question: I know that any universal object is initial/terminal in a suitable factor category, however, I can not make sense of this...

• I recall a "universal object"-style construction of the free group on $M$ in Algebra: Chapter 0 by Paolo Aluffi. As I recall, it's a pretty involved construction, requiring words, a reduction function on those words, and some other fancy footwork. To your second question, you should check out the concept of limits and Kan extensions, as those generalize a ton of different constructions in Category Theory. – Larry B. Oct 3 '18 at 20:02
• I am nowhere near an expert on this, but I see it this way:seemingly diverse mathematical constructions like direct products, tensor products, limits, colimits, kernels, etc all have the same idea in common: there is an initial (or terminal) object in a comma category and a "universal property" that characterizes how the constructions "work." Looked at this way, you can avoid working with some tedious constructions, like that of the tensor product , or even the Stone Cech compactification. Once you know these objects exist, you can work entirely with the universal property that represents them – Matematleta Oct 3 '18 at 20:18
• @LarryB. There is a general construction of a free object for an equational/algebraic theory by forming the term algebra and quotienting by the equations. This applies just fine to groups. There is no need for reduced words. The construction using reduced words is an alternative, isomorphic construction. The benefits of this more complex construction are lucidly explained in section 0.2 of these notes by Mike Shulman. – Derek Elkins left SE Oct 3 '18 at 20:35
• Very rough idea for (1) would be: show the subgroup $H$ of $F_M$ generated by the image of the map $M \to U(F_M)$ also satisfies the universal property. Thus, you get a unique isomorphism $H \simeq F_M$; and if you "unfold" the proofs enough, you can prove that one direction of this unique isomorphism ends up being the inclusion map $H \hookrightarrow F_M$. – Daniel Schepler Oct 3 '18 at 20:58
• It doesn't always make sense to think of a left adjoint as giving the smallest such-and-such. – Colin Oct 3 '18 at 21:16

These "smallest" intuitions are primarily coming from left adjoints of functors between posets. If $$f:A\to B$$ is a poset map, then if $$g$$ is left adjoint to $$f$$ we must have $$g(b)=\mathrm{inf} \{a:b\leq g(a)\}$$, by the definition of adjunctions. An instructive example is the case that $$A$$ is the poset of subgroups of a given group $$G$$, while $$B$$ is the poset of subsets, and $$f$$ sends a subgroup to its underlying set. Then $$f$$ has as a left adjoint the map $$g$$ which sends a subset $$S$$ to the subgroup generated by $$S$$. So this is completely concrete and proves a version of the intuitive picture you describe.

The rest of your question seems to conflate the subgroup generated by $$S$$ with the free group and the free abelian group generated by $$S$$, which are usually three wildly different objects. The free group is not the "smallest" group containing $$S$$ in the sense of the previous paragraph. Rather, it's the group with the least relations containing $$S$$. This kind of description of left adjoints is always correct in the context of algebraic structures, as long as the right adjoint forgets some structure. But it's not correct in general: for instance the discrete topological space on a set has the most open sets possible, so it's hard to view it analogously (although this can be done.)

In short, the way to express that the value of a left adjoint is the most general object with some property is by using the definition of adjunctions.

A free group on $$S$$ is the smallest group that contains the set $$S$$. How do we obtain the notion of “smallest” from the associated universal property?

Let's start at the beginning --a nice place to start.

Let's fake generality by calling the category of sets “simple” and the category of groups “furnished”.

Given a “simple” object $$S$$ --e.g., a set--, we want to construct the most general “furnished object” $$T$$ --e.g., a group-- that arises from $$S$$ by endowing it with the necessities to be considered a “furnished object”.

$$T$$ is known as the “free” furnished object on $$S$$; it is characterised by the universal property:

There's a simple map ι : S → T, such that for any simple map f : S → T'
there's a unique furnished map g : T → T' with f = g ∘ ι.


This expresses the fact that any furnished object $$T'$$ that contains a homomorphic image of $$S$$ will also contain a homomorphic image of $$T$$.

That is: $$T$$ is the most general furnished object containing a homomorphic image of $$S$$.

Rewriting the universal property symbolically:

$$∃ ι : S → T.\quad ∀ T'.\; f : S → T'\;.\; ∃₁ g : T → T'.\; f = g ∘ ι$$

Now foregoing the morphisms, and reading “→” as “≤”, leaves:

$$S ≤ T \quad∧\quad ∀ T'.\quad S ≤ T' \quad⇒\quad T ≤ T'$$

That is, $$T$$ is the least furnished object containing a homormorphic image of $$S$$.