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Recall that

  • The biinitial monoidal category with a monoid is given by the augmented simplex category together with the monoid $([0],\sigma^{0}_{0},\delta^{0}_{0})$ there.
  • The biinitial symmetric monoidal category with a commutative monoid is given by the pair $(\mathsf{FinSets},*)$ consisting of the category of finite sets and morphisms between them equipped with the coproduct as the monoidal structure, and the triple $(*,*\coprod*\to*,\emptyset\to*)$ with $*$ the punctual set as the commutative monoid.

There's a natural notion of a semiring object in a semiring category. Do we know what, if it exists, is the biinitial semiring category with a (commutative) semiring object?

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    $\begingroup$ Regarding the second bullet, I think you want to say that the initial symmetric monoidal category with a commutative monoid is given by the category of finite sets and all maps (not just bijections), with the cocartesian monoidal structure. The groupoid of finite sets is the initial symmetric monoidal category on no data. $\endgroup$
    – Tim Campion
    Commented Sep 2, 2021 at 14:16
  • $\begingroup$ Oh, right! Thanks, Tim! $\endgroup$
    – Emily
    Commented Sep 3, 2021 at 3:57
  • $\begingroup$ @TimCampion [Pedantic mode on] The initial symmetric monoidal category with no data is actually the contractible category. The groupoid of finite sets is the initial symmetric monoidal category on one object [Pedantic mode off] $\endgroup$
    – Denis Nardin
    Commented Sep 3, 2021 at 9:19
  • $\begingroup$ Oh geez of course, thanks! $\endgroup$
    – Tim Campion
    Commented Sep 3, 2021 at 12:56
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    $\begingroup$ I think that the category has objects the polynomial functors $\mathbf{FinSet} \to \mathbf{FinSet}$. The universal semiring object has underlying object $X$ (the identity), and every object has the form $\sum_i a_i X^{n_i}$. The tricky part is to work out the morphisms. For example, $\hom(X^n,X^m) = \hom_{\mathbf{FinOrd}}(n,m)$ and $\hom(n X,m X) = \hom_{\mathbf{FinSet}}(n,m)$, but $\hom(\sum_i a_i X^{n_i},\sum_j b_k X^{m_j})=?$. Probably it's a kind of semidirect product. $\endgroup$
    – Martin Brandenburg
    Commented Sep 3, 2021 at 18:28

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