Is a bialgebra pairing of Hopf algebras automatically a Hopf pairing?

Hi

I was in class when the question came up, and I remember the discussion on whether a bialgebra pairing of $A$ and $B$ always induces a bialgebra map $A\to B^\circ$. You guys tricked me back then :): it is, in fact, true. In other words, a bialgebra pairing is always (regardless of (non)degeneracy) exactly the same thing as a bialgebra map from one of the two bialgebras to the finite dual of the other.

I'm going to view the elements of $a$ as linear maps on $B$ via the pairing. We want to show that the map from $A$ to the dual of $B$ induced by the pairing actually lands in $B^\circ$.

This is pretty clear using the following characterization of the finite dual: $B^\circ$ is precisely the set of $f\in B^*$ for which one can find finitely many $g_i,h_i\in B^*$ such that $f(xy)=\sum_ig_i(x)h_i(y)$ (for all $x,y\in B$). It follows immediately from the bialgebra pairing conditions that any $a\in A$ satisfies this property: if, say $\Delta(a)=\sum_i a'_i\otimes a''_i$, then take $g_i=a'_i$ and $h_i=a''_i$.

Theo,

I think one can argue like this in the case of a non-degenerate pairing. I didn't check everything here carefully, so don't believe it unless you confirm it yourself.

One has from the pairing an inclusion of $i:X\hookrightarrow Y^*$. One has two maps on $i(X)$, $S_X$, and $S_Y^*|_{i(X)}$. Both of these satisfy the axioms of an antipode on $i(X)$ (which we can check by pairing with elements of $Y$), so they must agree, as desired, since being a Hopf algebra is a property, not a structure.

I'll think about the degenerate situation. My guess is that it's not true, but I don't know.