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

Denote category of monoids equipped with involution by $\textbf{invMon}$. Objects are pairs $\left(M,\iota\right)$ where $\iota$ is a map on the underlying set of $M$. Denoting $\iota$ by $x\mapsto\bar{x}$ we have $\bar{\bar{x}}=x$ and $\overline{x.y}=\bar{y}.\bar{x}$. Arrows in $\textbf{invMon}$ are homomorphisms that respect the involution. There is a forgetful functor $U:\textbf{invMon}\rightarrow\textbf{Set}$ and it has a left adjoint $F$. Every group equipped with the map $x\mapsto x^{-1}$ can be recognized as object of $\textbf{invMon}$ and every grouphomomorphism respects this involution. This gives a functor $I:\textbf{Grp}\rightarrow\textbf{invMon}$ and if $L$ is a left adjoint for $I$ then composite functor $LF:\textbf{Set}\rightarrow\textbf{Grp}$ should sends each set to a group free over the set. For object $\left(M,\iota\right)$ define $S=\left\{ x.\bar{x}\mid x\in M\right\}$ and $R=\left\{ 1\right\} \times S$. Let $C$ be the smallest congruence containing $R$ and let $M/C$ denote the 'quotient-monoid'. Then $\left[\bar{x}\right]\left[x\right]=\left[x\right]\left[\bar{x}\right]=\left[1\right]$ showing that $M/C$ is a group with $\left[\bar{x}\right]=\left[x\right]^{-1}$. So natural map $\nu:M\rightarrow M/C$ respects involution, hence is an arrow $\nu:\left(M,\iota\right)\rightarrow\left(M/C,inv\right)$ in $\textbf{invMon}$. Also $\left(M/C,\nu\right)$ is universal in the sence that for every arrow $\psi:\left(M,\iota\right)\rightarrow I\left(G\right)$ in $\textbf{invMon}$ there is a unique grouphomomorphism $\phi:M/C\rightarrow G$ with $\psi=I\varphi\circ\nu$. This means that $I$ indeed has a left adjoint. Then here we have the construction of a free group over a set that does not mention any equivalence classes of words or any reduced words. I am very suspicious, however. This because I never encountered this in literature.

My question is:

Am I overlooking something???

My second question is:

If it is oké then where can I find it in literature? I can't believe that it is new.

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If you construct the free group using the left adjoint of the forgetful functor $\mathbf{Grp} \to \mathbf{Set}$, you will also not need to mention any equivalence classes of words or reduced words... –  Zhen Lin Jun 2 '13 at 16:05
Mac Lane promotes the adjoint functor theorem in CWM by: "This left adjoint F:Set->Grp assigns to each set X the free group FX generated by X, so our theorem (=adjoint functor theorem) has produced this free group without entering into the usual (rather fussy) explicit construction of the elements of FX as equivalence classes of words in letters of X." Is that what you're talking about? –  drhab Jun 2 '13 at 18:20
How do you construct the left adjoint of $U$? –  Martin Brandenburg Jun 3 '13 at 8:14
set X goes to the monoid free over set X'=Xx{1}\/Xx{-1} (I mean a coproduct of X and X). This monoid has an involution sending word (x,1)(y,1)(z,-1) for instance to word (z,1)(y,-1)(x,-1). –  drhab Jun 3 '13 at 8:41
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