> So what you've constructed (I think!) as a morphism \\(\mathcal{X}\to\mathcal{Y}\\) in ME is a function \\(\phi\colon\text{Ob}(\\mathcal{X}) \to \text{Ob}(\\mathcal{Y})\\) and a monoid/group homomorphism \\(\hat\phi\colon \langle \mathcal{V}_X \rangle\to \langle \mathcal{U}_Y \rangle\\), such that \\(\hat\phi(\mathcal{X}(a,b)) = \mathcal{Y}(\phi(a),\phi(b))\\) .

That's right :D We have a functor.

The object mapping is \\(\langle\mathcal{V}_\bullet\rangle : \mathbb{ME} \to \mathbf{Grp}\\)

The morphism mapping is \\(\hat{\bullet} : \mathrm{hom}(\mathbb{ME}) \to \mathrm{hom}(\mathbf{Grp}) \\).

------------------------------------

Let \\(\mathcal{G}_G(g,h) = g^{-1}h \in G\\) be the enriched category John constructed in [#81](https://forum.azimuthproject.org/discussion/comment/18687/#Comment_18687)

Take have a group homomorphims \\(\psi : G \to H\\) for \\(G,H \in \mathbf{Grp}\\).

We can turn \\(\psi\\) into an \\(\mathbb{ME}\\) morphism with \\(\bar{\psi}\\) with

\\[
\bar{\psi} (g^{-1}h) = \psi(g^{-1})\psi(h)
\\]

I think \\(\langle\mathcal{V}_\bullet \rangle \\) is the left inverse of \\(\mathcal{G}•\\) up to isomorphism.