# What is the “on morphism” part of a free functor?

I’m trying to understand what a free functor is. In particular the free functor that is supposed to bring a set into the free monoid generated by it.

$F_X : \underline{Set} \to \underline{Mon}$

I understand that the “on object” part of the functor brings a set into the monoid like the following:

$F_X(X) = (\operatorname{List}(X), [], +)$

Where the right object is a monoid with the set of all possible lists composed by elements in $X$, the empty list as a neutral element, and a concatenation on lists as a operation.

It would seem to me that the “on morphism” part just turns a traditional function $f:X\to Y$ into monoid homomorphism in which the $\operatorname{List}(X)$ goes to a $\operatorname{List}(Y)$ in which you have all possible lists with the elements of the codomain of $f$.

If that’s the case in what sense is this functor free? It looks to me like a regular functor that just happens to point to a free object.

In short: a functor is free because it is defined by the universal property of the free object, not just because it happens to associate to each set its free object.

Let elaborate a litte more on this. Let start from the classical definiton of freeness in the context of (unary sorted set based) algebraic structure.

In this context we have a category of algebraic structures $\newcommand{\A}{\mathbb{A}}$ $\newcommand{\Set}{\mathbf{Set}}$ and a functor from such category to the category of sets, so we have a category $\A$ and the functor $U \colon \A \to \Set$.

A free object over a set $X \in \Set$ in nothing but an $A$-object $F(X)$ such that there is a function, i.e. a morphism of $\Set$, called $\eta_X \colon X \to U(F(X))$ with the property that every other map $f \colon X \to U(A)$ factors uniquely through an $A$-morphism $\bar f \colon F(X) \to A$.

Once one provides a choice of a such $A$-object $F(X)$ and a map $\eta_X \colon X \to U(F(X))$ satisfying the above universal property for each set $X$ by abstract nonsense you get an extension (necessarily unique) of the object-mapping $F$ to a functor which is a left adjoint ot $U$.

Of course different choices for the objects $F(X)$ or different choices for the arrows $\eta_X$ can produce different free functors, i.e. different adjoints for $U$. Luckily for us all these functors are naturally isomorphic.

Hope this helps.

"Free" doesn't refer to a property of functors that this one happens to have — it is a description of what the functor does.

• Does that imply that I got the morphism part right? Apr 18 '18 at 8:46
• @gurghet: I can't really make sense of your description on morphisms -- for $f : X \to Y$, what $F(f)$ does is take an element of $F(X)$ and rewrite it to become an element of $F(Y)$ by replacing each generator $x$ with the generator $f(y)$.
– user14972
Apr 18 '18 at 10:56

The free functor is the left adjoint to the forgetful functor. It is more then just a functor that sends an object to a free object.

So additional universal condition has to hold. In your case it can be stated as follows: if $G:Mon\to Set$ is the forgetful functor, $X\in Set$, $M\in Mon$ and $g:X\to G(M)$ is a function in $Set$ then it can be uniquely extended to a morphism $\overline{g}:F(X)\to M$.

It's analogous to how linear functions are fully determined by values on a linear base.

Also see wikpedia.

• How does $g$ look like though? $g(X)=\operatorname{List}(X)$? Apr 18 '18 at 8:48
• @gurghet You mean the forgetful functor? It's simply $G(X)=X$, $G(f)=f$. It just treats the monoid $X$ as a set and monoid morphism $f$ as a set function. Hence the name "forgetful". Apr 18 '18 at 8:54
• But the free monoid underlying set is $\operatorname{List}(X)$ not $X$ so I would be in a different place from where I started: $X\to \mathcal{M}\operatorname{List}(X) \to \operatorname{List}(X)$ Apr 18 '18 at 9:46
• @gurghet Yes, that's ok. You don't have to be in the same place. Read more about adjoint functors on wiki. Apr 18 '18 at 9:52