Zoran Skoda internal bialgebroid

This is an adaptation of a definition in

• Gabriella Böhm, Internal bialgebroids, entwining structures and corings, AMS Contemp. Math. 376 (2005) 207-226, math.QA/0311244

Let $C = (C,\otimes,1,\tau)$ be a symmetric monoidal category with symmetry $\tau$. Assume that $C$ allows coequalizers of parallel pairs which commute with $\otimes$.

Let $(R,m_R,\eta_R)$ be a monoid in $C$. An internal left $R$-bialgebroid in $C$ consists of the following data

• A monoid $(H,m_H,\eta_H)$ in $C$ equipped with two morphisms of monoids, $\alpha: R\to H$ and $\beta: R^{op}\to H$, such that $m_H\circ\tau_{H,H}\circ(\alpha\otimes\beta)=m_H\circ(\alpha\otimes\beta)$.

• Consider $H$ as an internal $R$-bimodule in $C$ via the left action $m_H\circ (\alpha\otimes H)$ and the right action $m_H\circ\tau_{H,H}\circ (H\otimes\beta)$. that $H\otimes H$ is a monoid in $C$ and the coequalizer $H\otimes_R H$. Denote by

  $$\pi:H\otimes H\to H\otimes_R H$$

  the canonical map of the coequalizer.

$$H\otimes R \otimes H\overset{H\otimes m_H\circ (\alpha\otimes H)}{\underset{m_H\circ\tau_{H,H}\circ (H\otimes\beta)\otimes H}{\Rightarrow}} H\otimes H\overset\pi\longrightarrow H\otimes_R H$$

The coequalizer is equipped with the unique right $H\otimes H$-action $\rho$ satisfying

$$\rho \circ (\pi \otimes \id_H \otimes \id_H) = \pi \circ m_{H\otimes H}$$

Require that

$$\rho\circ (\Delta\otimes\beta\otimes\eta_H) \cong \rho\circ (\Delta\otimes\eta_H\otimes\alpha),$$

where we identified the domains $H\otimes R\otimes 1$ and $H\otimes 1\otimes R$. In the case of vector spaces, this is the condition $h_{(1)}\beta(r)\otimes_R h_{(2)}=h_{(1)}\otimes_R h_{(2)}\alpha(r)$.

The above condition implies that there exist (unique) left action $\lambda : H\otimes (H\otimes_R H)\to (H\otimes_R H)$ such that

$$\lambda\circ(H\otimes\pi) = \rho\circ(\Delta\otimes H\otimes H)$$

The fact that $\lambda$ exists requires a long check in the categorical setup. It follows that $H\otimes_R H$ is an internal $H$-$H\otimes H$-bimodule in $C$.

• We require that ${}_R H_R$ be equipped with a comonoid structure $(H,\Delta,\epsilon)$ in the monoidal category of $R$-bimodules in $C$. Thus $\Delta:H\to H\otimes_R H$ is coassociative a map of $R$-bimodules, and $\epsilon : H\to R$ is a counit, also a map of $R$-bimodules. Require

$$\Delta\circ\eta_H = \pi\circ (\eta_H\otimes\eta_H)$$

In the case of vector spaces, this is the condition $\Delta(1) = 1\otimes_R 1$.

Now the most subtle axiom:

$$\Delta\circ m_H = \lambda\circ (H\otimes\Delta)$$

In the case when $C$ is the category of vector spaces, this is simply written $\Delta(h h') = h_{(1)} h'_{(1)}\otimes_R h_{(2)}h'_{(2)}$ for all $h,h'\in H$, but the fact that the right-hand side is well defined requires the axioms above.

A couple of axioms on $\epsilon$ are the remaining ones:

$$\epsilon\circ\eta_H = \eta_R$$

In the case of vector spaces, this is the condition $\epsilon(1_H) = 1_R$ (equivalently, the black action given by $h\blacktriangleright r = \epsilon(h\alpha(r))$ is unital).

$$\epsilon\circ m_H\circ(H\otimes\,\alpha\circ\epsilon) = \epsilon\circ m_H = \epsilon\circ m_H\circ(H\otimes\,\beta\circ\epsilon)$$

In the case of vector spaces, this is the condition $\epsilon(h\alpha(\epsilon(h')) = \epsilon(h h') = \epsilon(h\beta(\epsilon(h'))$ (or equivalently, the black action satisfies the action (associativity) axiom).