Isomorphism under restriction I need to show that $Hom_H(M|_H,\mathbb{C}_{triv})\cong Hom_G(M,Fun(G/H,\mathbb{C}))$, where $M$ is a representation of the finite group $G$, and where $H$ is a subgroup of $G$.
I have tried to relate $\mathbb{C}_{triv}$ to $(Fun(G,\mathbb{C})^H)$.
I then argue that, as we are restricting to $H$, we might as well write $Hom_G(M|_H,Fun(G,\mathbb{C})^H)$.
I know from a previous problem that this is isomorphic to $Hom_G(M|_H,Fun(G/H,\mathbb{C}))$ which is obviously not what I want.
I know that this is both hand-wavy and that all of the steps are questionable.
Is everything wrong? How should I tackle it?
I am not allowed to use character theory.
 A: $\DeclareMathOperator{\Fun}{Fun}
\newcommand{\C}{\mathbb C}$I assume you make the space of left cosets $G/H$ is a left $G$-module via
the usual formula $g(g'H) = gg'H$ so that $\Fun(G/H, k)$ is a left $G$-module via $(g \psi)(g'H) = \psi(g^{-1} g'H)$.
Given an $H$-linear map $\varphi : M \longrightarrow \C$, define a map $E_\varphi : M \longrightarrow \Fun(G/H,\C)$ so that
$$E_\varphi(m)(gH) = \varphi(g^{-1}m)$$
Notice that if $g_1H = g_2H$ then $g_1 = g_2h$ for some $h\in H$ and then
\begin{align*}
E_\varphi(m)(g_1H) &= \varphi(g_1^{-1}m) \\
&= \varphi(h^{-1}g_2^{-1}m) \\
&= h^{-1}\varphi(g_2^{-1}m) \\
&= \varphi(g_2^{-1}m) 
\end{align*}
since $\varphi$ is $H$-linear and the action on $\C$ is trivial. Thus $E_\varphi(m)$ is well defined. Moreover, for any $g,g_1\in G$:
$$E_\varphi(gm)(g_1H) = \varphi(g_1^{-1}gm) = E_\varphi(m)(g^{-1}g_1H) = (g\cdot E_\varphi(m))(g_1H)$$
so this map is $G$-linear. The inverse map sends a $G$-linear function $E : M \longrightarrow \Fun(G/H,\C)$ to the function $\varphi_E : M \longrightarrow \C$ such that  $\varphi_E(m) = E(m)(H)$.
