Can a free group over a set be constructed this way (without equivalence classes of words)?

Solution 1:

Looks fine. I haven't seen this construction before. Let me summarize it as follows:

We factor $\mathsf{Grp} \to \mathsf{Set}$ as $\mathsf{Grp} \to \mathsf{InvMon} \to \mathsf{InvSet} \to \mathsf{Set}$.

$\bullet$ The left adjoint of $\mathsf{InvSet} \to \mathsf{Set}$ sends a set $X$ to $X \times \{1\} \cup X \times \{-1\}$ with the involution $(x,1) \leftrightarrow (x,-1)$.

$\bullet$ The left adjoint of $\mathsf{InvMon} \to \mathsf{InvSet}$ sends a set $X$ with an involution $i$ to the free monoid on $X$ (the set of words over $X$ equipped with concatenation) with the involution $i(x_1 \dotsc x_n) = i(x_n) \dotsc i(x_1)$.

$\bullet$ The left adjoint of $\mathsf{Grp} \to \mathsf{InvMon}$ maps a monoid $M$ with an involution $i$ to the quotient $M/(x \cdot i(x)=1)_{x \in M}$.

It follows that $\mathsf{Grp} \to \mathsf{Set}$ has a left adjoint, namely the composition of the left adjoints $\mathsf{Set} \to \mathsf{InvSet} \to \mathsf{InvMon} \to \mathsf{Grp}$. This coincides with the usual construction (set of words $x_1^{\pm 1} \dotsc x_n^{\pm 1}$ modulo $x x^{-1} = 1$), but the factorization makes every step trivial, which is very nice.