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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?

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

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. 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.

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