One of the standard examples of a bicategory is the bicategory of rings (with bimodules as 1-morphisms), which is sometimes denoted $\operatorname{Bim}$ and in other sources $\operatorname{Ring}$ (or $\operatorname{Alg}$ if you are working over a commutative ground ring $k$). It is defined as follows:

  • objects are rings
  • the category of 1-morphisms from $R$ to $R'$ is the category of $R'$-$R$-bimodules (I'm working with left modules, but that doesn't really matter.)
  • composition is given by the tensor product of bimodules
  • 2-morphisms are bimodule homomorphisms
  • 2-composition is the usual composition of bimodule homomorphisms.

In my research, the following bicategory appeared "naturally" (I would like to call it $\operatorname{RingMod}$, resp. $\operatorname{AlgMod}$):

  • objects are pairs $(R,N)$ where $R$ is a ring and $N$ is an $R$-module
  • a $1$-morphism $(R,N)\to (R',N')$ is a pair $(M,g)$ where $M$ is an $R'$-$R$-bimodule and $g\colon M\otimes_R N\to N'$ is an $R'$-linear map
  • 1-composition is given by $(M',g')\circ (M,g)=(M'\otimes_{R'} M,g'(M'\otimes_{R'} g))$
  • 2-morphisms $\psi\colon (M,g)\to (\tilde{M},\tilde{g})$ are bimodule homomorphisms such that $\tilde{g} (\psi\otimes \operatorname{id}_N)=g$

I have two questions:

  1. [A bit vague] Does this seem like the correct notion of a 2-morphism in this bicategory or does it seem too strong?
  2. Did this bicategory already appear somewhere in the literature (possibly with a different notion of 2-morphism)?
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    $\begingroup$ Is that the slice 2-category over $\mathbf{Z}$? $\endgroup$ – Pavel Safronov Mar 25 '15 at 17:01
  • 2
    $\begingroup$ @PavelSafronov That's a nice observation. If I understand it correctly from the brief look I took at the definition it should be a lax slice $2$-category over $\mathbb{Z}$ (resp. $k$). $\endgroup$ – Julian Kuelshammer Mar 25 '15 at 17:09
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    $\begingroup$ Section 5.3 in arxiv.org/pdf/1106.5665.pdf looks very similar to what you have, but I am not sure if you have a special case they were using. $\endgroup$ – Aaron Chan Mar 31 '15 at 11:59

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