Question on Frobenius Reciprocity I have in my notes the statement of Frobenius reciprocity in the following two ways:

If $H\leq G$ and suppose that we have $\chi_1$ a character of $G$ and $\chi_2$ a character of $H$. Then:
  $$\langle \chi_1,\chi_2\uparrow^G\rangle_G=\langle\chi\downarrow_H,\chi_2\rangle_H$$

I also have it stated in the following way:

Suppose that we have $f:H\rightarrow GL(V)$ and $f':G\rightarrow GL(V)$ then we have that there is an isomorphism of the hom sets, that is:
  $$Hom_G(f',f\uparrow ^G)\cong Hom_H(f'\downarrow_H,f)$$

What I am failing to see is why these two statements are equivalent  (or really that related) 
Could somebody explain the correspondence here?
 A: $\def\Ind{\mathrm{Ind}}$
$\def\Hom{\mathrm{Hom}}$
Ok. So let's translate the statement about characters into a statement about irreducible representations.
First, note that both the Hom property, and the character inner product distribute over taking direct sums of representations. So, it suffices to show why
$$\langle \chi_{V}, \chi_{\Ind^{G}_{H} W} \rangle_{G} = \langle \chi_{V|_{H}}, \chi_{W}\rangle_{H}$$
and 
$$\Hom_{G}(V, \Ind^{G}_{H} W ) \cong \Hom_{H}(V|_{H}, W)$$
are equivalent for $V$ an irreducible representation of $G$ and $W$ an irreducible representation of $H$.
Now, in the case where the representations are irreducible, Schur's Lemmma tells us that
$$\Hom_{G}(V, \Ind^{G}_{H}W) \cong \Hom_{H}(V|_{H}, W)$$
says exactly that the number of times the irreducible $H$-representation $W$ appears in $V_{H}$ is the same as the number of times that the irreducible $G$ representation $V$ appears in $\Ind^{G}_{H}W.$
On the other hand, because characters of irreducible representations form an orthonormal basis for the inner product,
$$\langle \chi_{V}, \chi_{\Ind^{G}_{H} W}\rangle_{G}$$
is exactly the number of times $V$ appears in $\Ind^{G}_{H}W$ and 
$$\langle \chi_{V|_{H}}, \chi_{W}\rangle_{H}$$
is exactly the number of times $W$ appears in $V_{H}$. So the equality of the inner products also gives the exact same data that the equality of the $\Hom$'s did.
