How to define unitors of a category of enriched categories?

All categories in this question are assumed to be small. Let $$M$$ be a symmetric closed monoidal category. I will write the composition maps for $$M$$-(enriched) categories by $$\square_{a,b,c}:[a,b]\otimes[b,c]\to[a,c]$$.

For $$M$$-categories $$A,B$$, define an $$M$$-category $$A\otimes B$$ by

• $$\operatorname{Ob}(A\otimes B) = \operatorname{Ob}(A)\times\operatorname{Ob}(B)$$ (set-theoretic Cartesian product),
• $$[(a,b),(a'b')] = [a,a']\otimes[b,b']$$,
• $$\boldsymbol{1}_{(a,b)} = \left(i\to i\otimes i\xrightarrow{\boldsymbol{1}_a\otimes\boldsymbol{1}_b}[a,a]\otimes[b,b]\right)$$,
• $$\square_{(a,b),(a',b'),(a'',b'')} = \left(([a,a']\otimes[b,b'])\otimes([a',a'']\otimes[b',b''])\to([a,a']\otimes[a',a''])\otimes([b,b']\otimes[b',b''])\xrightarrow{\square\otimes\square}[a,a'']\otimes[b,b'']\right)$$.

For $$M$$-functors $$F:A\to C,\ G:B\to D$$, define an $$M$$-functor $$F\otimes G:A\otimes B\to C\otimes D$$ by

• $$(F\otimes G)(a,b) = (F(a),G(b))$$,
• $$(F\otimes G)_{(a,b),(a',b')} = F_{a,a'}\otimes G_{b,b'}$$.

I want to prove that MCat, the category of $$M$$-categories, is a monoidal category. For that, I should define suitable associators and unitors in MCat. I think I know how to define associators:

• $$\alpha_{A,B,C}(a,(b,c)) = ((a,b),c)$$,
• $$\left(\alpha_{A,B,C}\right)_{(a,(b,c)),(a',(b',c'))} = \alpha_{[a,a'],[b,b'],[c,c']}$$.

But I do not know how to define unitors. I think $$M$$ should be the unity object of MCat. If I am right, then for each $$M$$-category $$A$$, a left unitor $$\lambda_A$$ should send each $$(x,a)\in\operatorname{Ob}(M)\times\operatorname{Ob}(A)$$ to an object in $$A$$, and send each $$((x,a),(x',a'))$$ to a morphism $$[x,x']\otimes[a,a']\to[a,a']$$. But I cannot think of an obvious such assignment.

• No, the unit object $I$ of $M$-Cat is the category with a single object $*$, and where $I(*,*)$ is the monoidal unit of $M$. The unitors then are built out of the unitors of $M$, because $I\otimes A \cong A$ (obviously on objects, and $i\otimes A(x,y) \to A(x,y)$ is an isomorphism) Jun 23 at 9:47