Let $C$ be a coalgebra and $M$ a $C^*$-module, where $C^*$ is C's dual. Prove that $$\gamma: M \otimes C \rightarrow \text{Hom}_\mathbb{k}(C^*,M) \\ m \otimes c \mapsto [f \mapsto mf(c)]$$
is a $C^*$-module map.
In order to prove it I need to show that $$\gamma \circ \mu_{M \otimes C} = \mu_{\text{Hom}} \circ (Id_{C^*} \otimes \gamma)$$ where $\mu_{M \otimes C}$ is the action for the module $M \otimes C$ and $\mu_{\text{Hom}}$ is the action for $C^*$-module $\text{Hom}_\mathbb{k}(C^*,M)$, but I can't find how $\mu_{M \otimes C}$works.
$$\mu_{M \otimes C}: C^* \otimes M \otimes C \rightarrow M \otimes C \\ f \otimes m \otimes c \mapsto ? \\ \mu_{\text{Hom}}: C^* \otimes \text{Hom}_\mathbb{k}(C^*,M) \rightarrow \text{Hom}_\mathbb{k}(C^*,M) \\ f \otimes g \mapsto fg$$
Following the equality I should show that $\forall f \in C^*, \forall m \in M, \forall c \in C$ $$ (\gamma \circ \mu_{M \otimes C})(f \otimes m \otimes c)= (\mu_{\text{Hom}} \circ Id_{C^*} \otimes \gamma)(f \otimes m \otimes c) \\ \Rightarrow (\gamma \circ \mu_{M \otimes C})(f \otimes m \otimes c) = \mu_{\text{Hom}} (f \otimes \gamma(m \otimes c)) \\ \Rightarrow (\gamma \circ \mu_{M \otimes C})(f \otimes m \otimes c) = \mu_{\text{Hom}} (f \otimes m g(c)) \quad \forall g \in C^* \\ \Rightarrow (\gamma \circ \mu_{M \otimes C})(f \otimes m \otimes c) = mf g(c) \quad \forall g \in C^*$$
and given that I don't know how $\mu_{M \otimes C}$ works, I cant complete the equality.
Is it the rigth approach? Does anyone know how this multiplication acts on the elements of $C^* \otimes M \otimes C \rightarrow M \otimes C$?
Maybe there is a more subtle action of $C^*$ I'm not seeing, but it seems to me that the action of $C^*$ on $M\otimes C$ just comes from the action on $M$: in other words, $f\cdot (m\otimes c) = (f\cdot m)\otimes c$. Likewise, the action of $C^*$ on $\operatorname{Hom}_k(C^*,M)$ simply comes from the action of $C^*$, and $(f\cdot \varphi)(c)=f\cdot (\varphi(c))$.
In this case the proof that $\gamma$ is a $C^*$-morphism is just a one-line equation.