# From tensor algebras in monoidal categories to (commutative?) monoids

Let $D$ be a monoidal category (without the structure of a symmetric monoidal category), with unit object $Id$, and let $L$ be an invertible object in $D$, so that $L$ is dualizable and the pairing between $L$ and its dual $L^{*}$ is an isomorphism. (The situation I have in mind is that $D$ is the category of endofunctors of some category $C$ with composition giving the monoidal functor, and then $L$ is an autoequivalence of $C$.)

I want to put a monoid structure on $\coprod_{i \geq 0} D(Id, L^{\otimes i})$ by tensoring together two morphisms $\alpha: Id \rightarrow L^{\otimes i}$ and $\beta: Id \rightarrow L^{\otimes j}$ and then using the unit constraint on the domain $Id \otimes Id$ to get a morphism $\alpha \otimes \beta: Id \rightarrow L^{\otimes i +j }$.

Questions:

1. Does this construction make sense? I am a little worried that I have to fuss with the associativity constraints to really make sense of $L^{\otimes i}$ '.

2. Presuming that the construction makes sense, is the resulting product commutative? To try to check this, I would (and did) attempt the following: Given a product of two morphisms $\alpha \otimes \beta: Id \rightarrow L^{\otimes i} \otimes L^{\otimes j}$, tensor it on the right with $L^{\otimes -j}$ and on the left with $L^{\otimes j}$. This produces a morphism $L^{\otimes j} \otimes Id \otimes Id \otimes L^{\otimes -j} \rightarrow L^{\otimes j} \otimes L^{\otimes i} \otimes L^{\otimes j} \otimes L^{\otimes -j}$. Now use the unit and pairing on the domain and the pairing on the codomain to identify this with a morphism $Id \rightarrow L^{\otimes j} \otimes L^{\otimes i}$. Is this morphism $\beta \otimes \alpha$? And could this show that $\alpha \otimes \beta= \beta \otimes \alpha$, once the appropriate identifications of domain and codomain have been used?

• Probably your construction also works, if you replace $Id$ by a coassociative coalgebra in $D$. Sep 16, 2015 at 8:30
• The example you have in mind may be easier because a self-equivalence can be strictified to an automorphism, and then you can apply a non-linear version of Suarez-Alvarez, Mariano. “The Hilton-Heckmann Argument for the Anti-Commutativity of Cup Products.” Proceedings of the American Mathematical Society 132, no. 8 (2004): 2241–2246. doi.org/10.1090/S0002-9939-04-07409-X. Jan 2, 2019 at 22:21

EDIT: As Peter points out in the comments, my answer to number 2 is incorrect. When $$D$$ is braided, you can "pull out the twist" using naturality of the braiding, but for general monoidal categories, this morphism will not be equal to $$\beta \otimes \alpha$$.

However, there is an alternative procedure to the one you describe that always produces $$\beta \otimes \alpha$$, as shown in this even more crudely drawn diagram: This map corresponds to tensoring by $$L^{\otimes - j}$$ in the middle rather than on the right. In either case, $$\alpha \otimes \beta \neq \beta \otimes \alpha$$.

1. Yes, by coherence. For instance, if you take a strict monoidal category $$D'$$ equivalent to $$D$$, then any bracketing of $$L^{\otimes i}$$ in $$D$$ identifies with the canonical $$L^{\otimes i}$$ in $$D'$$ in a manner compatible with taking homs and tensor products. This guarantees that all ways you choose to bracket your $$L^{\otimes i}$$ and apply associators produce the same result, so your monoid is well-defined.
2. The result of your manipulations is $$\beta \otimes \alpha$$. This can be checked most easily using string diagrams. Here is my rather crude attempt to illustrate this: The $$\beta$$ can be "pulled around" to the left side of $$\alpha$$. However, this morphism is not the same as $$\alpha \otimes \beta$$ in general: this diagram is not isotopic to the one for $$\alpha \otimes \beta$$, so the result fails in the "free" case.

Note that the answers to both parts of your question are forms of coherence. The first is perhaps most easily seen as a case of "every monoidal category is equivalent to a strict one," while the second is most easily seen as a case of "isotopic string diagrams represent the same morphism."

• “The result of your manipulations is $\beta \otimes \alpha$” — are you sure this is right? From the string diagram we can see (by moving $\alpha$ over to the right) that it's $\beta' \otimes \alpha$, where $\beta'$ is “$\beta$ followed by a twist on $L^{\otimes i}$” (if we were in eg a tortile category, this would be literally true; in general, that's just what it kinda-sorta looks like), but surely in general the “twist” can be non-trivial? Aug 9, 2010 at 18:21

As Evan Jenkins already pointed out, we need a symmetry or at least a braiding for this to work. This is well-known, and I've also used this, but the first reference seems to be the quite recent preprint Coherence for invertible objects and multi-graded homotopy rings by Daniel Dugger.