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I am studying enriched categories, and as I wrote in my previous question How is the morphism of composition in the enriched category of modules constructed?, this is very difficult because there are no elementary books on this topic (so I hope that specialists in category theory will not kill me for what I am asking here).

In particular, I am looking from time to time at the Tannaka duality theorem for modules over monoids in a symmetric monoidal category, and I have a feeling that it can be proved without the Yoneda lemma which is suggested as an instrument in this situation.

Am I right?

My impression is that everything must be simple, although I can't restore the details (and I hope that somebody will help me).

Suppose we have a monoid $A$ in a symmetric monoidal category $V$ and we consider the enriched category $_AV$ of modules over $A$, and the forgetful functor $F:{_AV}\to V$. It's easy to see that the family of maps $$ \alpha_X:A\to[F(X),F(X)], $$ where $\alpha_X$ corresponds to the $A$-module structure $\mu_X:A\otimes X\to X$ on $X\in\operatorname{Ob}({_AV})$, is what is called a wedge (from $A$ to $[F,F]$, or what is the correct term?).

Our aim is to prove that for any other wedge $$ \beta_X:B\to[F(X),F(X)] $$ there is a unique morphism $\omega:B\to A$ such that the diagrams in the family universal property of wedges

(with $X\in\operatorname{Ob}({_AV})$) are commutative.

I guess that this $\omega$ is defined as the composition $$ B\xrightarrow{\beta_A}[A,A] \xrightarrow{[\iota,1_A]}[I,A] \xrightarrow{\rho_{[\iota,1_A]}^{-1}}[I,A]\otimes I \xrightarrow{\operatorname{ev}^A_I}A $$ (where $I$ is the unit in $V$, $\iota:I\to A$ the unit in the monoid $A$ and $\rho_{[\iota,1_A]}$ the right identity for $[\iota,1_A]$), but I don't understand how to prove the commutativity of diagrams (1).

In particular, the special case of (1) universal property of wedges with X = A

is also a puzzle for me. This is the same as the equality of the morphisms $$ B\otimes A\xrightarrow{\beta_A\otimes 1_A}[A,A]\otimes A \xrightarrow{[\iota,1_A]\otimes 1_A}[I,A]\otimes A \xrightarrow{\rho_{[\iota,1_A]}^{-1}\otimes 1_A}[I,A]\otimes I\otimes A \xrightarrow{\operatorname{ev}^A_I\otimes 1_A}A\otimes A \xrightarrow{\mu_A}A $$ and $$ B\otimes A\xrightarrow{\beta_A\otimes 1_A}[A,A]\otimes A \xrightarrow{\operatorname{ev}^A_A}A $$ but why are they equal? I can check this only in the situation when $A$ is a usual monoid (in the category of sets) or a usual algebra.

I would appreciate if anybody could show this "bookkeeping" (as Fosco Loregian calls it), because for an outsider like me this is incomprehensible. As I wrote in my previous question, each elementary step in this field is a problem, it requires weeks of pondering for me.

P.S. The applications of the Yoneda lemma are a gap in my education. If some kind person explains to me how it works, I will transfer this esoteric knowledge to the Moscow Mathematical School (with reference to this benefactor).

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    $\begingroup$ FWIW: Yoneda lemmas are really the fundamental theorems of category theory. Not using them is like not using the fundamental theorem of calculus to compute areas under a curve. $\endgroup$ Sep 30, 2021 at 14:35
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    $\begingroup$ (@LSpice Oh, I completely agree with your original comment. Indeed, common idioms are some of the most suspect parts of mathematical writing, and need to be used with very great care. Myself, I'm a big fan of Shakespearing puns, but would not use them in a main text, which I think of as like the primary lecture in a seminar. On the other hand, this comment section is more like the discussion over coffee or beer after the lecture: public, but still congenial.) $\endgroup$ Sep 30, 2021 at 14:46
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    $\begingroup$ Rather than me unpacking it, let me suggest a strategy. You know how to do the case when you have ordinary algebras or monoids. I assume this means: you know how to do the case when you have good access to "elements" of your algebras and modules. Probably every formula you write is linear in all of the elements that appear in it. If so, they you can usually implement those formulas internal to any enriching category by interpreting an expression like "for each $a \in A$" as "for each $a : A' \to A$", where not just $a$ but also $A'$ can vary. $\endgroup$ Sep 30, 2021 at 19:46
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    $\begingroup$ The advantage then is that the syntax of "$a \in A$" is left unchanged --- you don't have to change any formulas --- even though the semantics (the meaning) is different. But with the new semantics, the original formulas will probably supply a proof. This is a general strategy and I encourage you to try it in this case. $\endgroup$ Sep 30, 2021 at 19:48
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    $\begingroup$ Well, I don't want to promise that it will always work, but it is a very good strategy. $\endgroup$ Sep 30, 2021 at 22:36

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