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