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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 0x=00\cdot x = 0 follows from distributivity and additive inverses, since 0x+0x=(0+0)x=0x0\cdot x + 0\cdot x = (0+0)\cdot x = 0\cdot x and we can cancel one copy to obtain 0x=00\cdot x = 0. In a rig, however, we have to assert absorption separately. So shouldn’t a rig category also include isomorphisms x0x0xx\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 0x00 \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 xy+xzx(y+z)x \otimes y + x \otimes z \to x \otimes (y + z) is an isomorphism, then putting y=0y = 0, z=1z = 1 (the monoidal unit), we obtain a natural isomorphism

    x0+xxx \otimes 0 + x \to x

    whose restriction to the inclusion of xx is the identity xxx \to x. Let kk be its restriction to the inclusion of x0x \otimes 0. Then we obtain a bijection

    hom(x,x)k,idhom(x0,x)×hom(x,x)\hom(x, x) \stackrel{\langle 'k', id \rangle}{\to} \hom(x \otimes 0, x) \times \hom(x, x)

    and this forces hom(x0,x)\hom(x \otimes 0, x) to be a singleton, for any xx.

    • CommentRowNumber8.
    • CommentAuthorMike Shulman
    • CommentTimeJul 18th 2013

    Did you mean to say that we get a bijection

    hom(x,y)k,idhom(x0,y)×hom(x,y)\hom(x, y) \stackrel{\langle 'k', id \rangle}{\to} \hom(x \otimes 0, y) \times \hom(x, y)

    for any xx and yy, hence hom(x0,y)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=xy = x) since hom(x,x)\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.

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