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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 group homomorphism 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 group homomorphism $\phi:M/C\rightarrow G$ 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 okay then where can I find it in literature? I can't believe that it is new.

Shaun
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drhab
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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 02 '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 02 '13 at 18:20
  • How do you construct the left adjoint of $U$? – Martin Brandenburg Jun 03 '13 at 08:14
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    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 03 '13 at 08:41
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    My favorite construction of the free group: $F(S) := \langle \mathrm{im}(\phi) \rangle$ with $\phi : S \to \prod_{i : S \to |G|, ~ \langle \mathrm{im}(S) \rangle = G} G$, $\phi(s)_i=i(s)$. Notice that there is essentially only a set of groups in the indices. – Martin Brandenburg Jun 30 '14 at 20:39
  • @MartinBrandenburg Cool! It's so concise! Thank you! :) – Shaun Jul 01 '14 at 07:24
  • @MartinBrandenburg I've mentioned your comment in this question. I hope you don't mind :) – Shaun Jul 03 '14 at 20:39

1 Answers1

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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.

Martin Brandenburg
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  • Yes, that explains it very clearly. Thanks (again)! [I'll leave it for a couple of days before giving you the bounty - just to be fair! - but it's as good as yours :)] – Shaun Jul 01 '14 at 07:30
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    @Shaun I (just back from vacation) am pleasantly surprised by the fact that you have brought my question this way under attention again. The answer of Martin agrees with my own thinking about it. As he says: 'I haven't seen this construction before...'. Neither did I and that was my main motivation for asking. I think there is a way to generalize: invCat instead of invMon, having objects $(\mathcal C,\iota)$ where $\mathcal C$ is a category and $\iota$ a map on the arrows of $C$ sufficing $\iota\circ\iota=id$, $\iota(f\circ g)=\iota(g)\circ\iota(f)$ and $\iota(id_a)=id_a$. – drhab Jul 01 '14 at 08:20
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    Unfortunately I just don't know if this has already been written down somewhere. Of course it is elementary and nothing else than a (concise) reformulation of the usual construction, so that it is perfectly possible that someone has written it down. It is even more likely that many people have found this without making it public. – Martin Brandenburg Jul 01 '14 at 12:29
  • In the normal definition of the free group using words, the tedious part is associativity. Using this strategy for the proof, I feel like this complexity is pushed into showing that the "monoid quotient" by (x.i(x)=1) preserves the multiplication in the monoid (so that the quotient is still a monoid). Is there an easy way of proving this? – Vhailor Sep 24 '14 at 03:12
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    This is no issue. Given any relation on an algebraic structure, one considers the smallest congruence relation containing it. Then one takes the quotient and it is an algebraic structure of the same type again, for trivial reasons (essentially by the definition of "congruence relation"). – Martin Brandenburg Sep 24 '14 at 10:07