and only if $f$ induces the following commutative diagram for any $i,j\in G$: $$\xymatrix{
A_i\times M_j\ar[r]^(.6){{\rm act.}}\ar[d]^{1\times f}&M_{i+j}\ar[d]^f\\
A_i\times N_j\ar[r]^(.6){{\rm act.}}&N_{i+j}.}$$ This is equivalent to that the following diagram is commutative for any $h,h'\in H$: $$\xymatrix{
(A_{i-j+h})_{i,j\in I}\times(M_{i+h'})_{i\in I}\ar[r]^(.65){{\rm act.}}\ar[d]^{1\times f}&(M_{i+h+h'})_{i\in I}\ar[d]^f\\
(A_{i-j+h})_{i,j\in I}\times(N_{i+h'})_{i\in I}\ar[r]^(.65){{\rm act.}}&(N_{i+h+h'})_{i\in I}.}$$ This means that $f$ is a morphism in $\operatorname{\mathsf{Mod}}^HA^{[H]}$. We conclude that $F:\operatorname{\mathsf{Mod}}^GA\to\operatorname{\mathsf{Mod}}^HA^{[H]}$ is fully faithful.

Finally we show that $F$ is dense. For $i\in I$, let $e_i\in (A^{[H]})_0$ be an idempotent whose $(i,i)$-entry is $1$ and other entries are $0$. For any $N=\bigoplus_{h\in H}N_h$ in $\operatorname{\mathsf{Mod}}^HA^{[H]}$, let $$M_{i+h}:=e_iN_h\ \ \ \mbox{for $i\in I$, $h\in H$ and}\ \ \ M:=\bigoplus_{g\in G}M_g.$$ For $g,g'\in G$, we define the action $A_g\times M_{g'}\to M_{g+g'}$ by $$A_g\times M_{g'}=e_i(A^{[H]})_he_j\times e_jN_{h'}\xrightarrow{{\rm act.}} e_iN_{h+h'}=M_{g+g'},$$ where $i,j\in I$ and $h,h'\in H$ are unique elements satisfying $g'=j+h'$ and $g=i-j+h$. It is routine to check that $M$ is a $G$-graded $A$-module satisfying $FM\simeq N$.

It remains to show that $F$ induces an equivalence ${\operatorname{\mathsf{mod}}}^GA\to{\operatorname{\mathsf{mod}}}^HA^{[H]}$. This is immediate since $F(A(i+h))=(A^{[H]}(h))e_i$ holds for any $i\in I$ and $h\in H$.

Theorem \[graded Morita\] can also be shown by the following argument: Let $P:=\bigoplus_{i\in I}A(-i)\in\operatorname{\mathsf{Mod}}^GA$. Then the following statements can be checked easily:

-   ${\operatorname{End}\nolimits}_A^G(P)=A^{[H]}$.

-   $\operatorname{\mathsf{add}}\{P(h)\mid 