Let $H$ be a Hopf algebra, with invertible antipode, and let $(M,\Delta_M)$ and $(N,\Delta_N)$ be two left $H$-comodules. Now as we all know, we have a left $H$-comodule structure on the tensor product $M \otimes N$, defined by $$ (\Delta_M \otimes \Delta_N)(m \otimes n):= m_{(-1)}n_{(-1)} \otimes m_{(0)} \otimes n_{(0)}, ~~~~~ m \in M, n \in N. $$
Morover, we have canonical right $H$-comodule structures on $M$ and $N$ given by
$$
\Delta^R_M(m) := \tau \circ (S \otimes \text{id}) \circ \Delta_M(m) = m_{(0)} \otimes S(m_{(-1)}),
$$
and
$$
\Delta^R_N(n) := \tau \circ (S \otimes \text{id}) \circ \Delta_N(n) = n_{(0)} \otimes S(n_{(-1)}),
$$
where $\tau$ is the flip operator.
However, it seems to me that since $$ \tau \circ (S \otimes \text{id} \otimes \text{id}) \circ (\Delta_M \otimes \Delta_N)(m \otimes n):= m_{(0)} \otimes n_{(0)} \otimes S(m_{(-1)}n_{(-1)}) $$ while $$ (\Delta_M^R \otimes \Delta_N^R)(m \otimes n):= m_{(0)} \otimes n_{(0)} \otimes S(m_{(-1)})S(n_{(-1)}) = m_{(0)} \otimes n_{(0)} \otimes S(n_{(-1)}m_{(-1)}) $$ we do not in general have $$ \Delta_M^R \otimes \Delta_N^R = (\Delta_M \otimes \Delta_N)^R. $$ Is my reasoning correct here? If it is, then can anyone offer an explanation for why this should happen?
