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Take a ring $R$ and view it as being a category enriched over the category $\text{Ab}$ of abelian groups. Note: The Yoneda embedding $R \rightarrow \text{Fun} (R^{op}, \text{Ab})$ corresponds to the multiplication map $\mu : R \otimes R^{op} \rightarrow \text{Ab}$ under the yoneda embedding (edit: I meant under the hom-tensor adjunction for categories enriched over $\text{Ab}$). What is the coend of $\mu$ and what is the end of $\mu$? Also, is there a simple description of the twisted arrow category of $R \otimes R^{op}$?

I am trying to develop intuition for the general case.

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The sentence

The Yoneda embedding $R \rightarrow \operatorname{Fun} (R^{op}, \mathbf{Ab})$ corresponds to the multiplication map $\mu : R \otimes R^{op} \rightarrow \mathbf{Ab}$ under the Yoneda embedding.

is not quite correct ; instead I'd say that the Yoneda embedding $R \rightarrow \operatorname{Fun} (R^{op}, \mathbf{Ab})$ corresponds to the $\operatorname{Hom}$ functor $R \times R^{op} \rightarrow \mathbf{Ab}$ via "currying". Since the category $R\times R^{op}$ only has one object $*$, the arrow part of this functor then corresponds to the multiplication map $R\times R\to R$.

Now that we know this, we know the end of this functor $R \times R^{op} \rightarrow \mathbf{Ab}$ must be the abelian group $\operatorname{Nat}(Id_R,Id_R)$ of natural transformations of the identity endofunctor of $R$ to itself; it can be shown that this is isomorphic to the center $Z(R)$ of $R$. In fact we can see it rather directly : a wedge on this functor $R \times R^{op} \rightarrow \mathbf{Ab}$ is just an abelian group $A$ with one additive map $\varphi :A\to \operatorname{Hom}(*,*)=R$, such that for all $r\in R=\operatorname{Hom}(*,*)$ the square $$\require{AMScd}\begin{CD}A@>{\varphi}>> R \\ @V{\varphi}VV @VV{r\cdot (\_)}V \\ R @>>{(\_)\cdot r}> R \end{CD}$$ commutes, which is equivalent to asking that $r\varphi(a)=\varphi(a)r$ for all $a\in A$, is.e. that $\phi$ factors through $Z(R)$.

On the other hand a cowedge would be an abelian group $B$ and an additive map $\psi:R\to B$ such that $\psi(r\cdot s)=\psi(s\cdot r)$ for all $r,s\in R$, so the coend must be given by the universal such abelian group; this can be constructed as the quotient of $R$ by the subgroup generated by all terms of the form $r s-s r$, with the quotient map as $\psi$.

As for the twisted arrow category of $R$, it would have elements of $r$ as objects, and an arrow $r\to s$ would be a pair $(x,y)\in R^2$ such that $s=xry$. So it's exactly what you get if you see $R$ as an $R\otimes R^{op}$-module (i.e. a bimodule over itself) and apply the construction in this question!

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  • $\begingroup$ Thanks, this is a good solution. (When I said "under the yoneda embedding", I meant "under the hom-tensor adjunction for categories enriched over $\text{Ab}$", I just wasn't thinking straight). $\endgroup$ – Dean Young May 9 at 15:17

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