Short version: Let (C, ⊗, 1) be a locally presentable closed symmetric monoidal category with a zero object, and write 2 = 1 ∐ 1. Suppose the object 2 has a dual. Does it follow that C is a category with biproducts?

Longer version, with motivation: Let (C, ⊗, 1) be a locally presentable closed symmetric monoidal category. If you don't know what "locally presentable" means, you can replace these conditions with "complete and cocomplete symmetric monoidal category in which ⊗ commutes with colimits in each variable". Familiar examples include (Set, ×, •), (Set*, ∧, S0) (the category of pointed sets with the smash product), and (Ab, ⊗, ℤ). Any such category C has a unique "unit" functor FC : Set → C preserving colimits and the unit object: the set S is sent to the coproduct in C of S copies of 1. For a nonnegative integer n, let me also write n for the image under this functor of the n-element set. For instance, 0 represents the initial object of C.

A dual for an object X of C is another object X* together with maps 1 → X ⊗ X* and X* ⊗ X → 1 which satisfy the triangular identities; see wikipedia for more details. The data of X* together with these maps is unique up to unique isomorphism if it exists, so it makes sense to ask whether an object has a dual or not.

I'm interested in the relationship between which objects in the image of FC have duals and the existence of more familiar structures on C. In our examples,

  • C = Set: Only 1 has a dual.
  • C = Set*: Only 1 and 0 = • have duals.
  • C = Ab: n has a dual for any nonnegative integer n.

It's easy to show that 1 is always its own dual, and slightly less trivially, that 0 has a dual iff 0 is also a final object, i.e., C has a zero object, or equivalently C is enriched in Set*. Moreover, if C is semiadditive, i.e., enriched in commutative monoids, or equivalently has biproducts, then n has a dual (in fact, n is its own dual) for every nonnegative integer n. Conversely, if 0 has a dual, so that C is pointed, and 2 also has a dual, then there is a canonical map 2 = 1 ∐ 1 → 1 × 1 = 2*. My question, then, is: is this map is always an isomorphism? Or, could it happen that 2* exists but is not isomorphic to 2 via this map?

  • $\begingroup$ Why do you know $1\times 1=2^*$ when $2^*$ and $0^*$ exist? $\endgroup$ – Thomas Kragh Apr 6 '10 at 9:36
  • $\begingroup$ Because Hom(X, 2*) = Hom(2 ⊗ X, 1) = Hom(X ∐ X, 1) = Hom(X, 1) × Hom(X, 1) = Hom(X, 1 × 1). $\endgroup$ – Reid Barton Apr 6 '10 at 16:00

It is true that in a cosmos $\mathcal{V}$ (= a complete and cocomplete symmetric monoidal closed category) with zero object where $2$ has a dual, the canonical map $1+1 \rightarrow 1\times 1$ is invertible. In other words, you don't even need to assume that $\mathcal{V}$ is locally presentable. I think that this implies that the cosmos in question is semiadditive, which in turn implies that coproducts/products are biproducts (the only thing one needs to observe is that $C\otimes 1\times 1 \cong C\times C$, which follows from the fact that $1\times 1$ is the dual of $1+1$).

Proving this is a bit too involved for MO (because there is no good way of drawing string diagrams here). I have typed up an argument, which can be found here (Wayback Machine). I first prove that an autonomous symmetric monoidal category (autonomous means that all objects have duals) where the coproduct $1+1$ exists is semiadditive.

I can try to give some intuition for the argument in question. The key idea is that one wants to construct a diagonal map $1 \rightarrow 1+1$. The way to do this was inspired by the following paper:

André Joyal, Ross Street and Dominic Verity (1996). Traced monoidal categories. Mathematical Proceedings of the Cambridge Philosophical Society, 119 , pp 447-468 doi:10.1017/S0305004100074338

In my writeup, this "diagonal map" is the map $1\rightarrow 1+1$ in the string diagrams which is built out of a "loop" and the map $1+1 \rightarrow (1+1) \otimes (1+1)$ which I called $h$. From the formula given in the introduction of the paper by Joyal, Street and Verity it follows that my construction does indeed give the diagonal map in the case where $\mathcal{V}$ is the cosmos of vector spaces over some field.

Edit: updated expired link.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.