This is an adaptation of a definition in
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
the canonical map of the coequalizer.
The coequalizer is equipped with the unique right $H\otimes H$-action $\rho$ satisfying
Require that
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
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 $C$.
In the case of vector spaces, this is the condition $\Delta(1) = 1\otimes_R 1$.
Now the most subtle axiom:
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:
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).
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).
Last revised on March 25, 2015 at 17:43:45. See the history of this page for a list of all contributions to it.