TITLE: An elementary property of tensor products.
QUESTION [5 upvotes]: I'm studying representations theory to start my Masters thesis. I'm using the book of J. P. Serre, Linear Representations of Finite Groups and in the pg. 55 He affirm:

If $V$ is induced by $W$ and if $E$ is a $C[G]$-module, we have a canonical isomorphism 
  $$Hom^H(W,E)\simeq Hom^G(V,E)$$ 
  where $Hom^H(V,E)$ denotes the vector space of $C[G]$-homomorphisms of $V$ into $E$, and  $Hom^G(W,E)$ is defined similarly. This follows from an elementary property of tensor products.

Here, $G$ is a finite group, $V$ is a $C[G]$-module and $W$ a sub-$C[H]$-module of $V$.
I'm trying to prove this but without success. Can someone explain how to proceed here? What is this property and how to use it?
[EDIT]
In the book, He said to see also the following lemma:

Lemma: Suppose that $(V,\rho)$ is induced by $(W,\theta)$. Let $\rho':G\rightarrow GL(V')$ be a linear representation of $G$, and let $f:W\rightarrow V'$ be a linear maps such that $f(\theta_tw)=\rho'_tf(w)$ for all $t\in H$ and $w\in W$. Then there exists a unique linear map $F:V\rightarrow V'$ which extends $f$ and satisfies $F\circ \rho_s=\rho_s'\circ F$ for all $s\in G$

REPLY [4 votes]: This is a special case of the property called adjunction between the $Hom$ and the tensor functors.
Specifically, if $A$ and $B$ are two algebras over a commutative ring $k$, if $X$ is an $A$-$B$-bimodule, if $M$ is a right $A$-module and $N$ is a right $B$-module, then there is an isomorphism
$$ Hom_A(M, Hom_B(X, N)) \cong Hom_B(M\otimes_A X, N). $$
The property you wish to prove is then obtained by taking $A = C[H]$, $B=C[G]$, $M=W$, $N=E$ and $X=C[G]$.  Then we have $W\otimes C[H] = V$ and $Hom_{C[G]}(C[G], E) = E$, and thus obtain the desired isomorphism.
Now you can prove the adunction property directly, by defining the map
$$ Hom_B(M\otimes_A X, N) \to Hom_A(M, Hom_B(X, N)): f\mapsto (m\mapsto(x\mapsto f(m\otimes x))) $$
and showing that it is an isomorphism.