Let $\mathbf{Grpd}$ and $\mathbf{Cat}$ be respectively the 2-categories of small groupoids and of small categories. At the 1-categorical level, the inclusion $\mathbf{Grpd}\rightarrow\mathbf{Cat}$ has a right adjoint, namely the core. Thinking about the definition of the core of a category, I don't think that there is a way of extending it to natural transformations, essentially because if I have two functors $F, G: \mathcal{C}\rightarrow\mathcal{D}$ and a natural transformation $\alpha: F\Rightarrow G$, $core(\alpha) : core(F)\Rightarrow core(G)$ should be a natural isomorphism, and it is easy to find examples for which this cannot be true (one could be the determinant).
So my question is: does the inclusion $\mathbf{Grpd}\rightarrow\mathbf{Cat}$ have a right biadjoint? My impression is that the answer should be no, but I don't know how to prove this.
Best Answer
No, it doesn't. Bicategorical left adjoints preserve tensors with small categories. (If $x\in B$ is an object of a bicategory and $J$ is a category, the tensor $x\otimes J$ represents the pseudofunctor $y\mapsto B(x,y)^J$.) If $x$ is a groupoid and $[1]$ denotes the category freely generated by the graph $0\to 1$, then $x\otimes [1]$ is just $x\times I$, where $I$ is the groupoid freely generated by an isomorphism. But of course, after including $x$ into the bicategory of categories, $x\otimes [1]$ is simply $x\times [1]$, so the inclusion of groupoids into categories cannot be a left biadjoint.