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Is there a bicategory $V$ and a definition of monoid in a bicategory so that $\text{Monoids}(V)$ is the category of groups and homomorphisms?

EDIT: For example, is there a bicategory $V$ so that Monad(V) is the category of groups and group homomorphisms?

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    $\begingroup$ Welcome back, Joey! $\endgroup$ Mar 22, 2019 at 19:45

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There is a definition of monoids in a monoidal category and every monoidal category is a bicategory. You can try to extend this definition to a general bicategory. For example, we can say that a monoid in a bicategory $\mathcal{C}$ is a monoid in the monoidal category $\mathrm{Hom}_\mathcal{C}(X,X)$ for every object $X$ of $\mathcal{C}$ or that a monoid is a pair $(X,M)$, where $X$ is an object of $\mathcal{C}$ and $M$ is a monoid in $\mathrm{Hom}_\mathcal{C}(X,X)$. I'm not sure what are you looking for, so I can't say if these definition are what you need.

Anyway, any category $\mathcal{C}$ with finite coproducts is a monoidal category and the category of monoids in such a category is $\mathcal{C}$ itself. So, you can take $V$ to be the bicategory corresponding to the monoidal category of groups.

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    $\begingroup$ BTW, such an $(X,M)$ is usually called a monad in $\cal C$ (ncatlab.org/nlab/show/monad). $\endgroup$ Mar 22, 2019 at 19:06
  • $\begingroup$ Thanks Valery. Thanks Mike for the language I was looking for. What I’m looking for is a bicategory V so that Monad(V) is the category of Groups and group-homomorphisms. $\endgroup$
    – Joey Hirsh
    Mar 22, 2019 at 21:52
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    $\begingroup$ @JoeyHirsh The last paragraph answers your question. Take the bicategory with a single object $*$ and $\mathrm{Hom}(*,*)$ the category of groups. The composition is the coproduct of groups. Then the category of monads in this bicategory is equivalent to the category of groups. $\endgroup$ Mar 23, 2019 at 6:51
  • $\begingroup$ Oh! I misread that paragraph, thank you for telling me twice! $\endgroup$
    – Joey Hirsh
    Mar 23, 2019 at 17:59

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