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.



Show that a certain element is a linear combination of tensors
I think your argument shows that there exist $a,b,c$ with $$\Delta(a) = \sum\_i a^i\_{(1)} \otimes a^i\_{(2)}$$
$$\Delta(b) = \sum\_j b^j\_{(1)} \otimes b^j\_{(2)}$$


$$x \otimes y \otimes 1 = \sum\_i \sum\_j a\_{(1)}^i \otimes b\_{(1)}^i \otimes a\_{(2) }^ib\_{(2)}^jc $$


I don't really understand your notation but I think this is what you want to prove.


The proof you give is exactly correct, the point being that $y$ and $b$ don't depend on $i$ and thus it's not true that "some sums are depending on others".


A possible point of confusion is that the letter $b$ refers to two different things (it changes in the middle of the equation $y \otimes b = \sum b\_{(1)} \otimes b\_{(2)}c$).


