I posted this question on MSE but got no answer even after putting a bounty on it, so I figured I can try to ask here.
Let $(A, \Delta: A \to A \otimes A)$ be bialgebra (unital and counital) such that the map $$T: A \otimes A \to A \otimes A: a \otimes b \mapsto \Delta(a)(1 \otimes b)$$ is surjective.
We write $\Delta(a) = a_{(1)}\otimes a_{(2)}$ (Hopf-Sweedler notation). Is it true that we can write an element $x \otimes y \otimes 1$ as a linear combination of elements of the form $$a_{(1)}\otimes b_{(1)}\otimes a_{(2)}b_{(2)} c$$ where $a,b,c \in A$?
My attempt: I believe so. We can write $$x \otimes 1 = \sum a_{(1)}\otimes a_{(2)}b$$ (the summation depending on $a,b$). Hence, $$x \otimes y \otimes 1 = \sum a_{(1)}\otimes y \otimes a_{(2)}b= \sum (a_{(1)}\otimes 1 \otimes a_{(2)})(1 \otimes y \otimes b)$$ and we then write $y \otimes b = \sum b_{(1)}\otimes b_{(2)}c$ and we and up with something like $$x \otimes y \otimes 1 = \sum \sum a_{(1)}\otimes b_{(1)}\otimes a_{(2)}b_{(2)}c$$
I know I was not very careful with the summations, and some sums are depending on others, but I wanted to check if my idea was correct.