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(This is a cross-post from MathSE, as someone remarked that the question would be more appropriate on MO)

Fiore and Leinster have proved that if $\mathcal{A}$ is a monoidal category freely generated by one object $A$ and an isomorphism $\alpha: A \otimes A \to A$, then for every object $X \in \mathcal{A}, Aut(X)$ is isomorphic to the Thompson group $F$.

My question is the following: if we assume instead that $\alpha: A \otimes A \to A$ is not necessarily an isomorphism, and that there exist a morphism $\beta: A \to A \otimes A$ such that $\alpha \circ \beta = id$, is the result of Fiore and Leinster still true ?

I have a feeling we at least have $F \subset Aut(X)$. Loosely speaking, my approach is that since every element of $F$ can be represented as a pair $(R,S)$ of forests, we can always represent $R$ by a suitable composition of $\beta$ maps, then $S$ by a composition of $\alpha$ maps, the identity $\alpha \circ \beta = id$ ensuring that every facing caret gets cancelled to form a reduced forest diagram, i.e a unique element of $F$.

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In order to be precise, Fiore and Leinster do not work with existential quantifiers but with structures, i.e. they consider a free monoidal category on one object $A$ and an isomorphism $A\otimes A\rightarrow A$. The introduction, even the position of existential quantifiers may produce different answers to similar questions. – Fernando Muro Jun 27 '12 at 23:40
up vote 3 down vote accepted

Edit I noticed that in Fiore-Leinster preliminate the condition (free monoidal category of an isomorphism $ \alpha: A \otimes A \to A $) is different from what is written in the preliminary question, so I reworked my answer substantially.

In a Monoidal category $\mathcal{C}$ consider (a non empty) class of sections of the type $\beta: A\to A\otimes A$ and let $\Sigma$ its tensor product closure (finite tensor products of some morphisms of type $ \beta $ of the choose class and some identities).

From the article "Note on monoidal localizations " by B. Day (link text) the category of fraction $\mathcal{C}_\Sigma$ is (naturally) a monoidal category.

let $P: \mathcal{C}\to \mathcal{C}_\Sigma$ the natural functor.

The elements of $\Sigma$ are all monomorphisms (are sections) , and if $\Sigma$ admits a calculus of left fractions the canonical functors $P$ is faithful (see "Categories" H Shubert, 12.9.6(a), p.261). THen $\mathcal{C}$-$Aut(X)$ is a subgroup of $\mathcal{C}_\Sigma$-$Aut(X)$ (because $P$ is faithful).

Now consider the Monoidal category $[A, \alpha, \beta]$ free on (the condition):

"one object $A$ and on two morphisms $\alpha: A\otimes A\to A$, $\beta: A\to A\otimes A$, with $\alpha\circ \beta=1_A$".

This category has the following universal property: for any monoidal categories $\mathcal{C}$ with choose morphisms $a: X\otimes X\to X,\ b: X\to X\otimes X$ with $a\circ b=1_X$ there exists a unique strict monoidal functor $F_{a,b}: [A, \alpha, \beta]\to \mathcal{C}$ with $F(\alpha)=a,\ F(\beta)=b$.

Now in $[A, \alpha, \beta]$ consider the tensor closure $\Sigma$ of the section $\beta$,

and let $P:[A, \alpha, \beta]\to [A, \alpha, \beta]_\Sigma$ the category of fractions.

the category $[A, \alpha, \beta]_\Sigma $ has the universal property of the monoidal category on one isomorphisms

$\beta: A\to A\otimes A$ as in the FIore-Leinster article, then $F\cong [A, \alpha, \beta]_\Sigma$-$Aut(A)$.

Now, IF $\Sigma$ admit a calculus of left fraction then $P$ is faithful and $[A, \alpha, \beta]$-$Aut(A)$ is isomorphic to a subgroup of $F$.

P.S. I seems that $\Sigma$ admit a calculus of left fraction, but I have not checked it in detail

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Thank you for your answer, which is however very difficult for me to understand. When I posted the question, I had in mind a map $\alpha$ which would be surjective-like, thus I introduced a section $\beta$. But from your answer, why not use the localization of $\mathcal{A}$ at $\alpha$ instead of $\beta$ ? – AlexPof Jun 28 '12 at 7:49
Because in the theorme 12.9.6(a), p.261 of the H.Shubert Book, the mophisms of Sigma need to be monomorphisms. I improved my answere. – Buschi Sergio Jun 28 '12 at 15:34
You're right my question was badly formulated. I'll edit my question with the proper statement of Fiore and Leinster, and I'll add my comment in there... – AlexPof Jun 29 '12 at 8:20

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