I asked this a week ago at MSE, but without success.

I am studying enriched categories and I have a feeling that I am doing something wrong because all the way each step, each elementary proposition, requires enormous efforts and a lot of time from me. In particular, I am now stuck in the question of how the structure of the enriched category is introduced into the category $_AV$ of modules over a given monoid $A$ in a (symmetric) closed monoidal category $V$.

I asked this earlier (in April) at MSE:

if we consider an arbitrary closed monoidal category $V$ and take an arbitrary monoid $A$ in $V$, will the category $_A V$ of all left $A$-modules be an enriched category over $V$?

People explained to me the construction of the object of morphisms $V(M,N)$, I am grateful to them, and I accepted their answer, but not long ago I decided to write down the proof and (forgive me for my carelessness) I realized that I do not understand how the "enriched" composition morphism is constructed in $_AV$: $$ \circ_{L,M,N}:V(M,N)\otimes V(L,M)\to V(L,N). $$ I think that this must be the "lift to the equalizer" of the cone \begin{equation}\tag{1} V(M,N)\otimes V(L,M)\to \underline{\operatorname{Hom}}(M,N)\otimes \underline{\operatorname{Hom}}(L,M)\to \underline{\operatorname{Hom}}(L,N) \end{equation} for the pair of morphisms that define $V(L,N)$ as their equalizer: $$ \underline{\operatorname{Hom}}(L,N)\to \underline{\operatorname{Hom}}(A\otimes L,N) $$ and $$ \underline{\operatorname{Hom}}(L,N)\to \underline{\operatorname{Hom}}(L, \underline{\operatorname{Hom}}(A,N)) \to \underline{\operatorname{Hom}}(A\otimes L,N) $$ (here $\underline{\operatorname{Hom}}$ means the inner hom-functor in $V$). But the problem for me is that I don't understand why (1) is indeed a cone for this pair, i.e.

why the two morphisms $$ V(M,N)\otimes V(L,M)\to \underline{\operatorname{Hom}}(M,N)\otimes \underline{\operatorname{Hom}}(L,M)\to \underline{\operatorname{Hom}}(L,N)\to \underline{\operatorname{Hom}}(A\otimes L,N) $$ and $$ V(M,N)\otimes V(L,M)\to \underline{\operatorname{Hom}}(M,N)\otimes \underline{\operatorname{Hom}}(L,M)\to \underline{\operatorname{Hom}}(L,N)\to \underline{\operatorname{Hom}}(L, \underline{\operatorname{Hom}}(A,N)) \to \underline{\operatorname{Hom}}(A\otimes L,N) $$ coincide.

My proof requires 40 diagrams, but this is only a draft, because I deduce everything from the following lemma, which still is a puzzle for me:

Lemma.For each objects $A$, $X$, $Y$ and each morphism $\mu:A\otimes Y\to Y$ the following two morphisms coincide: $$ \underline{\operatorname{Hom}}(X,Y)\otimes A\otimes X\to \underline{\operatorname{Hom}}(X,\underline{\operatorname{Hom}}(A,Y))\otimes A\otimes X\to \underline{\operatorname{Hom}}(A\otimes X,Y)\otimes A\otimes X\to Y $$ (were the first arrow is generated by the arrow $Y\to\underline{\operatorname{Hom}}(A,Y)$, which corresponds to the morphism $\mu:A\otimes Y\to Y$), and $$ \underline{\operatorname{Hom}}(X,Y)\otimes A\otimes X\to A\otimes \underline{\operatorname{Hom}}(X,Y)\otimes X\to A\otimes Y\to Y $$ (were the last arrow is the morphism $\mu:A\otimes Y\to Y$).

This looks very strange to me (because it's difficult for me to believe that everything can be so complicated), but I don't see how this can be simplified, nor how to prove this lemma. Can anybody help me?

In Russia where I live, such problems are traditionally resolved by communication with experts, but as far as I know, there are no experts in category theory left in Russia.

isa V-category. 2. "there are no experts in category theory left in Russia" this reminds me of Hilbert's saying... $\endgroup$2more comments