## Background

For my PhD dissertation, I've developed a categorical generalization of many different systems of denotational semantics for light linear logic (LLL). I'd like to see if I can use this generalization to find a more "natural" (in the colloquial sense of the word) denotational semantics for LLL. At its core is a symmetric monoidal closed category with two functors on it. One of the functors is monoidal, and the other is not (well, it *could* be monoidal, but then it's a trivial example). There are some other requirements, but for the moment, I'm mostly curious about how common it is to have non-monoidal functors in the first place.

## Question

If you know of an example where someone uses a non-monoidal functor $T$ on a symmetric monoidal closed category $\mathbb{C}$, I'd like to hear about it. If you know of such a $T$ with natural transformations $d_A:TA \to TA\otimes TA$ and $e_A:TA \to 1$ forming comonoids for every object $A$, even better. If the category $\mathbb{C}$ also comes with a *monoidal* functor $S$, that would be even more fantastic. And if there's a natural transformation $T\Rightarrow S$, then I'll buy you dinner.

I've got examples (fibered phase spaces, stratified coherent spaces and locally bounded stratified cliques, games and discreet strategies, light length spaces), but they're all specifically created for this purpose, and I'm curious to see just how natural this kind of construction is.

isomorphisms$TA \otimes TB \to T(A \otimes B)$). Second, what do you mean by a functor "on" a category $C$? Do you mean an endofunctor of $C$, or a functor $C \to \text{Set}$, or just a functor with domain $C$? $\endgroup$ – Tom Leinster Mar 20 '12 at 14:02notbe cartesian. But what about $T(A)=$ the cofree comonoid on $A$? $\endgroup$ – Mike Shulman Mar 20 '12 at 20:58rightadjoint $!$. Then you get a comonad $T = U\circ !$ on $C$. We obviously have $d_A \colon TA \to TA \otimes TA, e_a \colon TA \to A$ as you require, and these being nat transformations to the comonoid morphism eqs for each $!f$. This looks like Lafont categories but without cocommutativity. However, I think that $T$ will be monoidal anyway; as ($C$ being symmetric) the left adjoint $U$ is strong monoidal and then by doctrinal adjunction $!$ is lax monoidal, so $T= U\circ !$ lax monoidal $\endgroup$ – Eduardo Pareja Tobes Mar 23 '12 at 15:384more comments