# Coend and End of this Functor

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.

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!

• 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). – Dean Young May 9 at 15:17