# Can you give an interesting example of useage of this Categorical definition of free object?

Definition 1.2 If $C$ is a concrete category, $F \in Obj(C)$, and $X$ is a set, and if $\phi : X \to \sigma(F)$ is a one-to-one function, then $F$ is called free on $X$ if and only if for every $A \in Obj(C)$ and set map $\psi : X \to \sigma(A)$, there exists a unique morphism $f \in Mor(F, A)$ such that $f \circ \phi = \psi$ as set maps from $X$ to $\sigma(A)$.

$\sigma$ is the map that comes together with the concrete category $C$, the map that makes $C$ into a concrete category.

Thanks. It's just that this definition is so abstract.

• Do you understand particular examples, when $C$ is, say, $\mathbf{Grp}$ and $\mathbf{CAlg}_k$. – Alex Youcis Nov 5 '13 at 0:10
• I might understand Grp. – BananaCats Category Theory App Nov 5 '13 at 0:11
• I mean, they're basically the objects that everything is a "quotient" of. Their use is, in some sense, undeniable. How often in field theory did you say "Ok, let $L/K$ be an extension, and let $\alpha\in L$. We know there is a $K$-algebra map $K[x]\to L$ with $x\mapsto \alpha$..." You are using PRECISELY that $K[x]$ is the "free commutative $K$-algebra on one generator". – Alex Youcis Nov 5 '13 at 0:14

Consider the category $\mathbf{\mathbb{K}\text{-}Vect}$ of $\mathbb{K}$-vector spaces. Let $V$ be a vector space, $B$ its basis and $\phi\colon B\to\sigma(V)$ the obvious embedding. Then $V$ is free on $B$, because the condition for freeness translates exactly to the property, that any linear map between vector spaces is completely determined by its action on the basis vectors.
A way to think about it : a free object on the set $X$ respects the universal property that $F(X)$ would respect if the forgetful functor $U \colon C \to \mathbf{Sets}$ had a left adjoint $F$.
So now, just think about what is a left adjoint of $U$ : it is a functor $F \colon \mathbf{Sets} \to C$ which comes with a (natural) bijection for all object $Y$ of $C$ $$\hom_{C}(F(X), Y) \simeq \hom_{\mathbf{Sets}}(X,U(Y)).$$ It means that defining a map from a free object on the set $X$ to the object $Y$ is the same as defining a map form $X$ to the underlying set of $Y$. Which is precisely what you do with the usual free object you (should) know : free groups, free modules, etc.