Question about functoriality of free group.

Question

This is an example from Riehl's category theory in context specifically 1.3.2.ix. She writes the following:

Example: There is a functor $F:\mathbf{Set} \rightarrow \mathbf{Group}$ that sends a set $X$ to the free group on $X$. This is the group whose elements are finite words whose letters are elements $x \in X$ or their formal inverses $x^{-1}$, modulo an equivalence relation that equates the words $xx^{-1}$ and $x^{-1}x$with the empty word. Multiplication is by concatenation with the empty word serving as the identity.

I'm familiar with the construction of the free group, but what I'm more puzzled by is this example doesn't specifically say anything about what this functor does to morphisms in the category $\mathbf{Set}$. That data seems to be required for this to be a functor, so for any set map $f:A \rightarrow B$ I would want some explicit description of a map $Ff:F(A) \rightarrow F(B)$.

I know that the free group has a universal property, in particular if $G$ is a group any set map $f:A \rightarrow G$ extends uniquely to a group homomorphism $\varphi:F(A) \rightarrow G$. I imagine that this property forces a unique association $f \mapsto Ff$ the functor requires, but I haven't been able to land on solid ground.

What I've been able to think of is the following. We know we have $\iota:B \hookrightarrow F(B)$ as a set map, and so their composition is another set map $\iota f:A \rightarrow F(B)$. Then the universal property of the free group tells us there is a unique group hom $\varphi:F(A) \rightarrow F(B)$ for which $\varphi|_A = \iota f$. But this doesn't seem to force a unique group hom $F(A) \rightarrow F(B)$ for each $f:A \rightarrow B$, just a unique one that restricts to $\iota f$.

I guess I still don't feel that I have what I want: For any $f:A \rightarrow B$ an explicit definition of the associated map $Ff:F(A) \rightarrow F(B)$. Please let me know if I can clarify and thanks in advance.

Edit: I think my issue can be formulated succinctly as the following: Because the other didn't explicitly say what this functor does to morphisms in the category of sets, I assume there must be only one way for a functor that associates a set to it's free group to associate a set map to a group homomorphism. I don't see how what I have justifies the uniqueness of this group hom. in that sense. Maybe my assumption is wrong however.

Answer 1

Your issue appears to be:

But this doesn't seem to force a unique group homomorphism $()\to ()$ for each [map] $:\to $, just a unique one that restricts to $\iota $.

Nobody says that there is a unique group homomorphism $()\to ()$. The claim is that there is a unique group homomorphism $h_f: ()\to ()$ extending $\iota $, which is something you already know. This is the functoriality statement that you are looking for: For some reason you are refusing to accept it as such.

NB. The map $h_f$ clearly satisfies:

• $h_{f_1\circ f_2}= h_{f_1}\circ h_{f_2}$.

• $h_{\mathrm{id}_A}= \mathrm{id}_{F(A)}$.

which is one of the functoriality requirements.

Answer 2

I guess I still don't feel that I have what I want: For any $f:A \rightarrow B$ an explicit definition of the associated map $Ff:F(A) \rightarrow F(B)$. Please let me know if I can clarify and thanks in advance.

I think the ambivalence comes about because you haven't actually written down an explicit construction for the map witnessing the universal property.

Write $G:\mathbf{Group} \to \mathbf{Set}$ for the functor that forgets group structure. If we have a map of sets $f: A \to GB$, and the free group construction $FA$ is given by "words" as in your text, then how are we to construct the corresponding map of groups $f^{\mathsf{T}}: FA \to B$?

If it isn't given in your reference: the answer is to "substitute letters in place". For instance, if $A=\{a,a'\}$, and $f: A\to GB$ is given by $$\begin{align} f(a)&=\beta&f(a')&=\beta' \end{align}$$ then $f^{\mathsf{T}}:FA \to B$ sends the four-letter word $[a',a,a',a^{-1}]: FA$ to $$f^{\mathsf{T}}([a',a,a',a^{-1}])=\beta'\beta\beta'\beta^{-1}\text{.}$$ The technical part is in actually defining $f^{\mathsf{T}}$ by induction on the size of words and showing that it respects equivalence classes.