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I don't have access to the 1980 paper that allegedly proves that there are only two symmetric monoidal closed structures on Cat, so I tried a baby version of the problem, reconstructing the proof on

For every object in

(and perhaps a few graph homomorphisms between them) to understand the action of ⊗ on the whole category.

Also, I know ⊗ has an identity object, I. Can I be the empty graph? No, because of preservation of colimits, G ⊗ 0 = 0, so that invalidates I being the identity if ever G is not zero. So now consider what I could possibly be. Whatever it is, it's the colimit of a bunch of copies of

a)

b) Either

Consider the case where I has no arrows. It's a collection of n vertices. But

The case where

**Grph**. It's tougher than I thought to get all the way there, but I see a few inroads. It's impressive how strong the assumption is latent in the word "closed". To assume, as is a consequence of assuming monoidal-closedness, that A ⊗ — preserves colimits gives you a lot!For every object in

**Grph**can be gotten by taking some suitable colimit of the one-vertex graph**1**and the one-edge graph**2**. This means that I really only need to understand**1**⊗**1**,**1**⊗**2**,**2**⊗**2**(and perhaps a few graph homomorphisms between them) to understand the action of ⊗ on the whole category.

Also, I know ⊗ has an identity object, I. Can I be the empty graph? No, because of preservation of colimits, G ⊗ 0 = 0, so that invalidates I being the identity if ever G is not zero. So now consider what I could possibly be. Whatever it is, it's the colimit of a bunch of copies of

**1**and**2**. And observe that**1**⊗ I =**1**doesn't have any arrows in it. But again since ⊗ preserves colimits,**1**⊗ I has to be the colimit of a bunch of copies of**1**⊗**1**and**1**⊗**2**. Weeelll, maybe there are no copies of**1**⊗**2**in there, since we haven't shown that I has any arrows, but at least there's a copy of**1**⊗**1**. Since colimits never get rid of arrows, only identify them with other arrows, we conclude thata)

**1**⊗**1**has no arrows in it.b) Either

**1**⊗**2**has no arrows, or I has no arrows (which would cause copies of**1**⊗**2**to show up in the colimit that's equal to**1**⊗ I =**1**)Consider the case where I has no arrows. It's a collection of n vertices. But

**1**= I ⊗**1**= n copies of**1**, so n = 1, and**1**= I. Therefore also we know the value of**1**⊗**2**: it's**2**. Down this road is the non-cartesian tensor product that I described yesterday. All that remains is to somehow prove that**2**⊗**2**is a four-arrow square.The case where

**1**⊗**2**has no arrows ought to lead inexorably to the cartesian product, but I haven't made any further progress on that front.