TITLE: Details of Defining Free Group in Terms of Universal Property
QUESTION [6 upvotes]: The only definition of the free group I have seen thus far (undergraduate) is an explicit construction
via reduced words and concatenation.
What I am now trying to do is see if it is possible to define the free group abstractly in terms of the universal property. The definition I think should work is:
A free group $F_S$ on a set $S$ is a group generated by $S$ such that for any group $G$ and any function $f: S \rightarrow G$, $f$ extends to a unique homomorphism $\tilde{f}: F_S \rightarrow G$.
So what I want to do now is prove existence and uniqueness (up to isomorphism?) from this abstract definition, and I'm struggling to make progress, so I'm looking for tips, specifically:

Is my definition correct, or is there something I'm missing?
Any tips on how to make progress on existence?
Any tips on how to make progress on uniqueness?

REPLY [3 votes]: Your definition is right and uniqueness follows from it from a purely categorical argument. Namely, you're defining a free group $F_S$ over $S$ and, to make things more precise, you should make an "inclusion" $\iota:S \to F_S$ part of the definition, so that for every group $G$ and every function $f:S \to G$, there is a unique homomorphism $\widetilde{f}:F_S\to G$ with $f = \tilde{f} \circ \iota$.
Assume that $F_{S,1}$ and $F_{S,2}$ are two free groups over $S$, with inclusions $\iota_i:S \to F_{S,i}$, $i=1,2$. Apply the definition of free group to $\iota_1: S \to F_{S,1}$ to get a homomorphism $\widetilde{\iota_1}:F_{S,2}\to F_{S,1}$ with $\iota_1 = \widetilde{\iota_1}\circ \iota_2$. Switching roles of $F_{S,1}$ and $F_{S,2}$, we also obtain a group homomorphism $\widetilde{\iota_2}:F_{S,1}\to F_{S,2}$ with $\iota_2= \widetilde{\iota_2}\circ \iota_1$. Now we have a homomorphism $\widetilde{\iota_1}\circ\widetilde{\iota_2}:F_{S,1}\to F_{S,1}$, which has to be the identity, as $$(\widetilde{\iota_1}\circ\widetilde{\iota_2})\circ \iota_1 = \widetilde{\iota_1}\circ (\widetilde{\iota_2}\circ \iota_1) = \widetilde{\iota_1}\circ \iota_2 = \iota_1 $$and $${\rm Id}_{F_{S,1}}\circ \iota_1 = \iota_1.$$More precisely, we're able to conclude that $\widetilde{\iota_1}\circ\widetilde{\iota_2} = {\rm Id}_{F_{S,1}}$ because of the uniqueness of extensions in the definition. Switching roles of $F_{S,1}$ and $F_{S,2}$ again, we also get that $\widetilde{\iota_2}\circ\widetilde{\iota_1} = {\rm Id}_{F_{S,2}}$. So $F_{S,1}\cong F_{S,2}$.
Here's what's going on behind the scenes: defining a category $\mathcal{C}_S$ by setting $${\rm Obj}(\mathcal{C}_S) = \{ (F,\iota) \mid \iota:S \to F \mbox{ and $F$ is a group}\}$$ and $${\rm Mor}_{\mathcal{C}_S}((F_1,\iota_1),(F_2,\iota_2)) = \{ \phi:F_1\to F_2 \mid \phi \mbox{ is a homomorphism and }\phi\circ \iota_1=\iota_2\},$$a free group $(F_S,\iota)$ over $S$ is nothing more than an initial object in $\mathcal{C}_S$. But it's a general fact (which I invite you to prove --- mimicking the above) that in any category, initial objects are unique up to (unique!) isomorphism.
However, all of this fluff does not establish existence of $F_S$. But since we already know that if it exists, it will be unique up to isomorphism, to get existence it suffices to exhibit one construction. Then the usual construction fits the bill.
The universal property gives an elegant way of constructing homomorphisms between free groups. For example, if $S_1$ and $S_2$ are sets and we have a function $f:S_1\to S_2$, then apply the universal property of $F_{S_1}$ to $\iota_2\circ f$ to get a group homomorphism $\widetilde{\iota_2\circ f}:F_{S_1}\to F_{S_2}$. If $f$ is bijective, then $\widetilde{\iota_2 \circ f}$ is an isomorphism with inverse $\widetilde{\iota_1\circ f^{-1}}$ (check this without picking elements, it's instructive).