# Right (bi)adjoint of the inclusion of $\mathbf{Grpd}$ in $\mathbf{Cat}$

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.

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.