User alexpof - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:48:46Z http://mathoverflow.net/feeds/user/20903 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100802/thompsons-group-f-and-monoidal-categories Thompson's group F and monoidal categories AlexPof 2012-06-27T19:56:43Z 2012-06-29T09:16:37Z <p>(This is a cross-post from MathSE, as someone remarked that the question would be more appropriate on MO)</p> <p><a href="http://arxiv.org/pdf/math/0508617v2.pdf" rel="nofollow">Fiore and Leinster</a> 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 $F$.</p> <p>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 ? </p> <p>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 $F$.</p> http://mathoverflow.net/questions/100802/thompsons-group-f-and-monoidal-categories/100810#100810 Comment by AlexPof AlexPof 2012-06-29T08:20:34Z 2012-06-29T08:20:34Z 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... http://mathoverflow.net/questions/100802/thompsons-group-f-and-monoidal-categories/100810#100810 Comment by AlexPof AlexPof 2012-06-28T07:49:38Z 2012-06-28T07:49:38Z 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 $\beta$ ?