Let $H$ be an Hopf Algebra, $M,N$ $H$-modules on the left, then we can define in $Hom(M,N)$ a natural structure of $H-$module (on the left) given by $(h*f)(m) = h_{(1)}f(S(h_{(2)})m), \forall m \in M$, where $S$ is the antipode of $H$.
Now show that $Hom_H(M,N) = Hom(M,N)^H$, where $Hom(M,N)^H =\{f \in Hom(M,N): h*f = \epsilon(h)f, \forall h \in H\}$.
My effort:
Let $f \in Hom_H(M,N)$ then $(h*f)(m) = h_{(1)}f(S(h_{(2)})m)$, since $f$ is $H$ linear we have that $h_{(1)}f(S(h_{(2)})m) = h_{(1)}S(h_{(2)})f(m)$, and $h_{(1)}S(h_{(2)}) = \epsilon(h)f(m)$. So we have
$(h*f)(m)= \epsilon(h)f(m)$, for all $m \in M$. That is, $f \in Hom(M,N)^H$.
For otherwise, if $f \in Hom(M,N)^H$ then we have, for all $h \in H$, $h_{(1)}f(S(h_{(2)})m) = \epsilon(h)f(m)$. But $e(h) = h_{(1)}S(h_{(2)})$, so, if $h_{(1)} \neq 0$ we have $f(S(h_{(2)})m) = S(h_{(2)}) f(m)$.
My question is, how to do when $h$ is not in image of $S$??
To prove the other inclusion, we just need to establish $H$-linearity, i.e. $f(hm) = h f(m)$. Thus we start by writing $f(h m)$ as $f(\epsilon(h_{(1)})h_{(2)}m)$. Pulling $\epsilon(h_{(1)})$ out, we get $\epsilon(h_{(1)}) f(h_{(2)}m)$. Next we use the fact that $f\in \operatorname{Hom}(M, N)^H$, so that $\epsilon(h_{(1)}) f(h_{(2)}m) = h_{(1)} f(S(h_{(2)})h_{(3)}m)$. Finally we can collapse the factors $S (h_{(2)}) h_{(3)}$ to $\epsilon(h_{(2)})$ and pull it out of $f$, so that this is equal to $h_{(1)}\epsilon(h_{(2)}) f(m) = h f(m)$.