Does the 'arrow' internal category forgetful functor have a right adjoint?

Let $$\mathcal{E}$$ be a category with finite limits, and let $$Arr : Cat(\mathcal{E}) \to \mathcal{E}$$ be the forgetful functor that sends an internal category to its object of arrows/morphisms. Does this functor always have a right adjoint? In particular, does $$Arr : Cat \to Set$$ have a right adjoint?

One conjecture I have is that if $$\mathcal{E}$$ also has binary coproducts and $$X$$ is an object of $$\mathcal{E}$$, then the right adjoint to $$Arr$$ would send $$X$$ to the indiscrete internal category on the $$\mathcal{E}$$-object $$X + X$$ of objects.

I am particularly interested in knowing whether $$Arr : Cat(\mathcal{E}) \to \mathcal{E}$$ preserves (binary) coproducts in general, which is true in the case $$\mathcal{E} = Set$$.

• I suppose a related question would be: if $\mathcal{E}$ has binary coproducts, then does $Cat(\mathcal{E})$ have them as well, and do we know how they are constructed? Jan 22, 2021 at 16:43

No, $$Arr$$ does not even have a right adjoint in the case $$\mathcal{E}=\mathsf{Set}$$. The reason is that it fails to preserve coequalizers.
Any monoid $$M$$ can be viewed as a category with one object (so the monoid is the set of arrows under composition). The coequalizer of a pair of monoid homomorphisms in the category $$\mathsf{Mon}$$ of monoids agrees with the coequalizer of the corresponding pair of functors in $$\mathsf{Cat}$$. But the forgetful functor from $$\mathsf{Mon}$$ to $$\mathsf{Set}$$ does not preserve coequalizers, so neither does $$Arr$$.