Suppose $k$ has characteristic not two and $A=k\langle x,y:x^2=1, y^2=0\rangle$ with $\Delta(x)=x\otimes x$, $\Delta(y)=y\otimes 1+x\otimes y$, $\varepsilon(x)=1$ and $\varepsilon(y)=0$; this is the Sweedler Hopf algebra. Let $E$ be the subHopf algebra generated by $x$, which has $\{1,x\}$ as a basis. Then $R=k\otimes_EH$ R=k\otimes_EA$has$\{\overline y=1\otimes 1=1\otimes 1,\overline y=1\otimes y\}$as a basis, and its coalgebra structure is given by$\Delta(\overline 1)=\overline 1\otimes\overline 1$,$\Delta(\overline y)=\overline y\otimes\overline1+\overline1\otimes\overline y$,$\varepsilon(\overline1)=1$and$\varepsilon(\overline y)=0$. Since$E\cong k\times k$as an algebra, the category$\mathrm{Mod}_E$is semisimple. On the other hand, suppose$M\in\mathrm{Mod}_A^R$. One can check that the right$R$-comodule structure$\rho$of$M$is determined by a linear map$\phi:M\to M$such that$\phi^2=0$by the equation $$\rho(m)=m\otimes\overline 1+\phi(m)\otimes\overline y.$$ Similarly, the$A$-module structure on$M$is easily seen to be such that$m\cdot y=0$for all$m\in M$and$\phi(m\cdot x)=\phi(m)\cdot x$for all$m\in M$. It follows that one can identify an object$M$of$\mathrm{Mod}_A^R$with a$4$-tuple$(M^+,M^-,\phi^+,\phi^-)$such that$M=M^+\oplus M^-$is the decomposition of$M$as direct sum of the eigenspaces of right multiplication by$x$(the only possible eigenvalues are$1$and$-1$, and it is diagonalizable) and$\phi^{\pm}:M^\pm\to M^\pm$are the restrictions of the map$\phi$to$M^+$and$M^-$(so in particular they square to zero). Moreover, morphisms in$\mathrm{Mod}_A^R$have the obvious description in terms of these$4$-tuples. Now, it is very easy to see using this description that$\mathrm{Mod}_A^R$is not semisimple: for example, the object$(k^2,0,\left(\begin{array}{cc}0&1\\0&0\end{array}\right),0)$is not semisimple (in fact, the category is the direct sum of two copies of the module category over the quiver$\bullet\to\bullet$). It follows that$\mathrm{Mod}_E$and$\mathrm{Mod}_A^R$are not equivalent in this case. (The answer is yes, though, in the two extreme cases where (i)$E=k$or (ii)$E=A$(the first one is the «fundamental theorem of Hopf algebras», the second one is trivial) 1 A very small example where the answer is no: Suppose$k$has characteristic not two and$A=k\langle x,y:x^2=1, y^2=0\rangle$with$\Delta(x)=x\otimes x$,$\Delta(y)=y\otimes 1+x\otimes y$,$\varepsilon(x)=1$and$\varepsilon(y)=0$; this is the Sweedler Hopf algebra. Let$E$be the subHopf algebra generated by$x$, which has$\{1,x\}$as a basis. Then$R=k\otimes_EH$has$\{\overline y=1\otimes 1,\overline y=1\otimes y\}$as a basis, and its coalgebra structure is given by$\Delta(\overline 1)=\overline 1\otimes\overline 1$,$\Delta(\overline y)=\overline y\otimes\overline1+\overline1\otimes\overline y$,$\varepsilon(\overline1)=1$and$\varepsilon(\overline y)=0$. Since$E\cong k\times k$as an algebra, the category$\mathrm{Mod}_E$is semisimple. On the other hand, suppose$M\in\mathrm{Mod}_A^R$. One can check that the right$R$-comodule structure$\rho$of$M$is determined by a linear map$\phi:M\to M$such that$\phi^2=0$by the equation $$\rho(m)=m\otimes\overline 1+\phi(m)\otimes\overline y.$$ Similarly, the$A$-module structure on$M$is easily seen to be such that$m\cdot y=0$for all$m\in M$and$\phi(m\cdot x)=\phi(m)\cdot x$for all$m\in M$. It follows that one can identify an object$M$of$\mathrm{Mod}_A^R$with a$4$-tuple$(M^+,M^-,\phi^+,\phi^-)$such that$M=M^+\oplus M^-$is the decomposition of$M$as direct sum of the eigenspaces of right multiplication by$x$(the only possible eigenvalues are$1$and$-1$, and it is diagonalizable) and$\phi^{\pm}:M^\pm\to M^\pm$are the restrictions of the map$\phi$to$M^+$and$M^-$(so in particular they square to zero). Moreover, morphisms in$\mathrm{Mod}_A^R$have the obvious description in terms of these$4$-tuples. Now, it is very easy to see using this description that$\mathrm{Mod}_A^R$is not semisimple: for example, the object$(k^2,0,\left(\begin{array}{cc}0&1\\0&0\end{array}\right),0)$is not semisimple (in fact, the category is the direct sum of two copies of the module category over the quiver$\bullet\to\bullet$). It follows that$\mathrm{Mod}_E$and$\mathrm{Mod}_A^R$are not equivalent in this case. (The answer is yes, though, in the two extreme cases where (i)$E=k$or (ii)$E=A\$ (the first one is the «fundamental theorem of Hopf algebras», the second one is trivial)