turning left modules into right modules over bialgebroids Let $\mathcal{B}$ be a left R-bialgebroid as defined in https://arxiv.org/pdf/1403.3597.pdf on page 87. Let $M$ be a left $\mathcal{B}$-module. Can $M$ be made a right $\mathcal{B}$-module (as for the Hopf algebras, it can be done using the antipode)?
 A: I think that the question is not completely well-posed, but let me try to give an answer anyway.
In general, no, you cannot: there is no "natural" way to convert left modules into right modules. A first vague idea of why this is the case is given by the facts that any ordinary bialgebra over a field $\Bbbk$ is a particular example of a $\Bbbk$-bialgebroid (and we know that for general bialgebras is not necessarily true) and that to speak about modules you are only using the algebra structure of the bialgebroid, whence you are essentially asking if given an algebra over a commutative ring you may convert actions on one side into actions on the other.
The key fact when doing this over a Hopf algebra is the fact that the antipode is an anti-homomorphism of algebras, but you indeed don't need really an antipode to convert left (resp. right) modules into right (resp. left) ones: an antimultiplicative and unital map would be more than enough.
So, let me sum up a bit. In general, there is no natural way to switch the side of an action, however


*

*You can, of course, define a new action on the right and make of a left module a bimodule (and hence, in particular, a right module), but this is not necessarily connected with the original left action.

*If your bialgebroid $B$ (and in general any algebra) is endowed with an anti-multiplicative and unital endomorphism $f:B^{op}\to B$, you can use the restriction of scalars to convert any left $B$-module into a left $B^{op}$-module (i.e. a right $B$-module) and viceversa.

*If your bialgebroid $B$ (and in general any algebra) is commutative, then $\mathsf{Id}_B$ is an anti-multiplicative and unital endomorphism and you may apply the previous construction.


This is of course not exhaustive at all, but I hope it answers your question.
