1
$\begingroup$

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.

$\endgroup$
0
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Does that imply that I got the morphism part right? $\endgroup$
    – gurghet
    Apr 18 '18 at 8:46
  • 1
    $\begingroup$ @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 $f(y)$. $\endgroup$
    – user14972
    Apr 18 '18 at 10:56
0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ How does $g$ look like though? $g(X)=\operatorname{List}(X)$? $\endgroup$
    – gurghet
    Apr 18 '18 at 8:48
  • $\begingroup$ @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". $\endgroup$
    – freakish
    Apr 18 '18 at 8:54
  • $\begingroup$ But the free monoid underlying set is $\operatorname{List}(X)$ not $X$ so I would be in a different place from where I started: $X\to \mathcal{M}\operatorname{List}(X) \to \operatorname{List}(X)$ $\endgroup$
    – gurghet
    Apr 18 '18 at 9:46
  • $\begingroup$ @gurghet Yes, that's ok. You don't have to be in the same place. Read more about adjoint functors on wiki. $\endgroup$
    – freakish
    Apr 18 '18 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.