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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeSep 10th 2012

concerning the discussion here: notice that an entry rig category had once been created, already.

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeJul 17th 2013

In a ring, the absorption/annihilation law $0\cdot x = 0$ follows from distributivity and additive inverses, since $0\cdot x + 0\cdot x = (0+0)\cdot x = 0\cdot x$ and we can cancel one copy to obtain $0\cdot x = 0$. In a rig, however, we have to assert absorption separately. So shouldn’t a rig category also include isomorphisms $x\otimes 0 \cong x \cong 0\otimes x$ as part of its structure?

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeJul 17th 2013

(And indeed, Laplaza includes them.)

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeJul 17th 2013

It’s worth noting though that in the case where $\oplus$ is the categorical coproduct, that $0 \otimes x \cong 0$ comes for free.

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeJul 17th 2013

Are you saying that if we assume $\otimes$ preserves binary coproducts in each variable, then it automatically preserves initial objects in each variable as well?

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeJul 17th 2013

Also: changed.

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeJul 17th 2013

Yes: if we assume the natural canonical map $x \otimes y + x \otimes z \to x \otimes (y + z)$ is an isomorphism, then putting $y = 0$, $z = 1$ (the monoidal unit), we obtain a natural isomorphism

$x \otimes 0 + x \to x$

whose restriction to the inclusion of $x$ is the identity $x \to x$. Let $k$ be its restriction to the inclusion of $x \otimes 0$. Then we obtain a bijection

$\hom(x, x) \stackrel{\langle 'k', id \rangle}{\to} \hom(x \otimes 0, x) \times \hom(x, x)$

and this forces $\hom(x \otimes 0, x)$ to be a singleton, for any $x$.

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeJul 18th 2013

Did you mean to say that we get a bijection

$\hom(x, y) \stackrel{\langle 'k', id \rangle}{\to} \hom(x \otimes 0, y) \times \hom(x, y)$

for any $x$ and $y$, hence $hom(x\otimes 0,y)$ is a singleton? Anyway, this is nice — it should go at distributive monoidal category.

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeJul 18th 2013

I thought my reasoning worked (where $y = x$) since $\hom(x, x)$ is inhabited. No?

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeJul 19th 2013
• (edited Jul 19th 2013)

My apologies: I see what Mike is saying, and now I don’t see how to prove my claim. (What happened is that I remembered a (true) statement that the claim holds in the case of cartesian monoidal categories, and thought that would generalize right away to the more general case.)

(And, in fact, it’s trivially false. For example, the coproduct $\vee$ on a join-semilattice preserves binary coproducts in each variable but not the initial object.)

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeJul 19th 2013

I added to distributive monoidal category a modified claim (remark 1) and proof.

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeJul 20th 2013

Okay, great. I moved this remark out of the “Definition” section where I didn’t think it exactly belonged.