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Take an arbitrary category $C$, and consider $End(C)$, any endofunctor category consisting of some or all endofunctors of $C$. I have several related questions:

  1. What restrictions must we impose on the base category $C$ such that $End(C)$ has a symmetric monoidal product?
  2. What restrictions must we impose on the functors $F \in Ob(End(C))$ for $End(C)$ to have a symmetric monoidal product?

I am also interested in the same two questions when we are interested in dagger compactness, ie, there is an endo functor $\dagger : End(C) \rightarrow End(C)$, where $\dagger$ is identity on objects but flips all the natural transformations in $End(C)$, ie all its arrows. $\dagger \circ \dagger = Id_{End(C)}$.

I am trying to build Abramsky, Coecke's semantic category for quantum protocols (also Selinger's construction) in terms endofunctor categories on arbitrary base categories.

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  • $\begingroup$ I would like to point out that with a dagger structure on $End(C)$, we have a category where every object is a monad and has a twin comanad, which together form a Frobenius monad. $\endgroup$
    – Ben Sprott
    Aug 2, 2014 at 17:58
  • $\begingroup$ Sorry, not every object is a monad, but when you do have a monad, you also have a frobenius monad. $\endgroup$
    – Ben Sprott
    Aug 2, 2014 at 18:43
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    $\begingroup$ I'm not sure about the dagger part, but any time you have a category of functors Fun(C,D) you can endow it with the Day convolution product. In your case C=D and you don't need any restrictions on the types of functors. You just need C to be a monoidal category (symmetric if you want the Day product symmetric). See ncatlab.org/nlab/show/Day+convolution, mathoverflow.net/questions/130616/… $\endgroup$
    – David White
    Aug 3, 2014 at 0:04
  • $\begingroup$ @DavidWhite: Day convolution needs colimits with distribute over the tensor product. $\endgroup$
    – Martin Brandenburg
    Aug 14, 2014 at 12:56
  • $\begingroup$ Hi Martin. Thanks. I was unaware of that, since I always work in a closed monoidal setting. But it's good to know there is in fact an extra hypothesis needed. $\endgroup$
    – David White
    Aug 14, 2014 at 14:12

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