Assume that $B$ is an algebra which is also a coalgebras (we do not assume that $B$ is a bialgebra: we do not assume $\Delta(cd)=\Delta(c)\Delta(d)$). Assume that $A$ is $B$-comodule algebra. Then we have a map $\delta: A \to B \otimes A$ which satisfies $(1 \otimes \delta)\delta = (\Delta \otimes 1)\delta$, where $\Delta: B \to B \otimes B$ is the comultiplication on $B$. Suppose that for all $a,b \in A$, we have $\delta(ab)=\delta(a)\delta(b)$. Do we have $\Delta(cd)=\Delta(c)\Delta(d)$ for all $c, d \in B$?
Let $a,b \in A$ and $\delta(a)=a_{(1)} \otimes a_{(2)}$, $\delta(b)=b_{(1)} \otimes b_{(2)}$. Then \begin{align} (1 \otimes \delta)\delta(ab) & = (1 \otimes \delta)(\delta(a)\delta(b)) \\ & = (1 \otimes \delta)(a_{(1)}b_{(1)} \otimes a_{(2)} b_{(2)}) \\ & = a_{(1)}b_{(1)} \otimes \delta(a_{(2)}b_{(2)}) \\ & = a_{(1)}b_{(1)} \otimes \delta(a_{(2)})\delta(b_{(2)}) \\ & = (a_{(1)} \otimes \delta(b_{(1)}))(a_{(2)} \otimes \delta(b_{(2)})) \\ & = ((1 \otimes \delta)\delta(a))((1 \otimes \delta)\delta(b)) \\ & = ((\Delta \otimes 1)\delta(a))((\Delta \otimes 1)\delta(b)) \\ & = (\Delta(a_{(1)}) \otimes a_{(2)})(\Delta(b_{(1)}) \otimes b_{(2)}) \\ & = ( \Delta(a_{(1)}) \Delta(b_{(1)}) ) \otimes a_{(2)}b_{(2)}. \quad (1) \end{align} On the other hand, \begin{align} (1 \otimes \delta)\delta(ab) & = (\Delta \otimes 1)\delta(ab) \\ & = (\Delta \otimes 1)(\delta(a)\delta(b)) \\ & = (\Delta \otimes 1)(a_{(1)}b_{(1)} \otimes a_{(2)} b_{(2)}) \\ & = \Delta(a_{(1)}b_{(1)}) \otimes a_{(2)} b_{(2)}. \quad (2) \end{align} By $(1)$ and $(2)$, do we have $\Delta(cd)=\Delta(c)\Delta(d)$ for all $c,d \in B$? Thank you very much.
Edit: assume that $B$ is an algebra which is also a coalgebras (but not necessarily bialgebra). Assume that $A$ is $B$-comodule algebra.