Words in the Category of Sets

Question

I was wondering about free objects in different categories and the "words" in those categories. I think I have a generally good grasp on the idea, but I started to think about stranger free objects such as free objects in the category of Sets, Rings, Fields, and Modules, as well as hypothetical structures with ternary+ operations. I understand the universal property of free groups, and to some extent of arbitrary categories, but I cannot fully comprehend the analogue of "words" in other free objects. Free Sets would be strange, what would words of free sets look like if that even could make sense? I was thinking it might just be the set of all singleton sets of each element in your generating set, so that the only words are the actual elements of the set. Or it might be useful to define it as the empty set? I have no idea. Any insight is wholly welcome.

Answer 1

If $C$ is a concrete category i.e. a category equipped with a faithful functor $U : C \to \mathsf{Set}$, then the free $C$-object over a set $X$ is by definition an object $F(X)$ which satisfies the universal property $\hom(F(X),A) \cong \hom(X,U(A))$, naturally in $A \in C$. If $C=\mathsf{Set}$ and we choose (what else?) $U=\mathrm{id}$, then clearly we have $F(X)=X$ for every $X$. The free set on a set $X$ is just $X$. You can also view it as the coproduct (here: disjoint union) of free sets on one generator, and the free set on one generator is the one-element set. Every $x \in X$ is also a word in the free set on $X$, but there is no possibility (or need) to combine any two words.

If we consider the category of pointed sets $\mathsf{Set}_*$ with the usual forgetful functor $(X,*) \mapsto X$, then the free pointed set on a set $X$ turns out to be $X \sqcup \{*\}$ (we adjoin formally a new point).

Here is a more interesting example: A magma is just a set together with a binary operation (no rules such as associativity are required). We have the concrete category of magmas. The free magma on a set $X$ consists of words such as $((x_1 x_2) x_3) (x_4 x_5)$. You can view these words as finite binary trees whose leaves are labelled with elements from $X$.

       .
      / \
     /   \
    /\   /\
   / x3 x4 x5
  /\    
 /  \   
x1  x2

The free $R$-module on a set $X$ consists of formal linear combinations $\sum_{x \in X} \lambda_x \cdot x$, where $\lambda_x \in R$, almost all $\lambda_x=0$ so that this is a finite sum. The free $R$-algebra on a set $X$ (where $R$ is a commutative ring) is the polynomial algebra $R[\{T_i\}_{i \in X}]$ whose indeterminates are indexed by $X$.

There are no free fields on a set, even not on the empty set, since there is no initial field. Rather, for every prime number $p$ resp. $p=0$, there is an initial field of characteristic $p$, namely $\mathbb{F}_p$ resp. $\mathbb{Q}$. The free ring of characteristic $p$ on one generator is $\mathbb{F}_p[T]$, but in order to get a free field we would have to mod out some irreducible polynomial, but no such polynomial is universal. We could also take the field of fractions $\mathbb{F}_p(T)$, but this is the free field of characteristic $p$ on one trancendental generator.

Notice that the definition of free objects also works if $C$ is not a category of algebraic structures. Consider for example $C=\mathsf{Top}$, the category of topological spaces. Then $F(X)$ is easily seen to be $(X,\wp(X))$, i.e. the set $X$ equipped with the discrete topology.