> 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.