Isomorphic representations, Schur's lemma I'am supposed to show the following:

Let $G_1,G_2$ be finite groups and $\rho_1: G_1\rightarrow GL(V_1), \space \rho_2:G_2 \rightarrow GL(V_2)$ representations of the mentioned groups. The representations $\rho_1, \rho_2$ induce a representation (via tensor product) $\rho_1 \otimes\rho_2$ of $G_1 \times G_2$ on $V_1 \otimes V_2$ in the following way:
  $$(\rho_1 \otimes \rho_2)(g_1,g_2)=\rho_1 (g_1) \otimes \rho_2(g_2).$$ 
Let $\Gamma:G_1\times G_2 \rightarrow V_1\otimes V_2$ be a irreducible representation of $G_1\times G_2$. Show that $\Gamma$ is isomorphic to $\rho_1 \otimes \rho_2$ for some irreducible representations $\rho_1$ and $\rho_2$. 

Schur's lemma : (one of the many versions)
Let $\rho_1,\rho_2:G \rightarrow Aut(V_1), Aut(V_2)$ be two irreducible representations. Let $f: V_1 \rightarrow V_2 $ be a $G$-equivariant map. Then the following hold for $f$:
   1. $f$ is a null map. 
   2. $f$ is an isomorphism and in $V_1,V_2$ to $f$ corresponds a (quadratic) matrix $R$, such that: $\rho_2(g)=R \rho_1(g) R^{-1}, \forall g\in G$. We say that $\rho_1$ and $\rho_2$ are isomorphic. 
My try: 
I was thinking in terms of the above (Schur's) lemma. We know that if $\rho_1$ and $\rho_2$ are irreducible, then $\rho_1 \otimes \rho_2$ is also irreducible. Therefore we have two irreducible representations, namely:
$$\rho_1\otimes\rho_2: G_1 \times G_2 \rightarrow V_1 \otimes V_2$$
$$\text{and}$$
$$\Gamma: G_1\times G_2 \rightarrow V_1 \otimes V_2 .$$
Additionally let's choose:
$$f:V_1\times V_2 \rightarrow V_1 \times V_2: f=id \text{    (identity map)}$$
Since both $\rho_1\otimes\rho_2$ and $\Gamma$ are irreducible and $f$ is a $(G_1\times G_2)$-equivariant map, according to the above lemma we have two possibilities: 


*

*$f$ is a null map. 


This is clearly not the case since $f=id$
and then we have:


*$f$ is an isomorphism and in $V_1,V_2$ to $f$ corresponds a (quadratic) matrix $R$, such that: $\Gamma(g)=R (\rho_1 \otimes \rho_2)(g) R^{-1}, \forall g\in G_1 \times G_2$. We say that $\Gamma$ and $\rho_1 \otimes \rho_2$ are isomorphic. 


This clearly is the case since $f=id$ is a isomorphism. And therefore we have shown that $\Gamma$ and $\rho_1 \otimes \rho_2$ are  isomorphic for some irreducible $\rho_1$ and $\rho_2$.

Question:
I would appreciate some input here, I also know that there are many ways to prove this. This is either a realy (stupidly) simple question or I have grossly miscalculated myself here. 
Also see:
Related question
 A: This does not work : why is $f$ equivariant ? 
Given representations $\rho_1,\rho_2$ of $G$ on $V$, $id_V: (V,\rho_1)\to (V,\rho_2)$ is equivariant if and only if $\rho_1=\rho_2$ : this has nothing to do with Schur's lemma, or irreducibility, this is actually very easy to see. 
Now you claim that $id$ is equivariant but to prove that you'd have to first prove that $\Gamma = \rho_1\otimes \rho_2$, which is what you want to prove anyway. 
Note that your proof can't work anyway because you chose arbitrary $\rho_1,\rho_2$, so it's clear that it can't work. 
Martin Brandenburg's answer in your related question proves this fact (he mentions it in his answer) in the case of an algebraically closed field of characteristic prime to $|G_1|,|G_2|$; if you can prove that $\rho_1\otimes \rho_2\cong \rho'_1\otimes \rho'_2 \implies \rho_1\cong \rho'_1 \land \rho_2\cong \rho'_2$. 
However, Mariano Suarez-Alvarez already gives a proof (still in your related question) of your statement : he actually produces two irreducible representations $V_1,V_2$ of $G_1,G_2$ respectively whose tensor product is isomorphic to the $\Gamma$ you start with - you should definitely check out his answer, it's well-written !
The only point in it that is not entirely clear at a first glance is why $\hom_G(U,V_{\mid G}) $ has dimension $\leq \frac{\dim V}{\dim U}$, but that follows at once by using Schur's lemma and decomposing $V_{\mid G}$ into irreducible representations 
(although to make sure that that holds you probably have to use some hypothesis, again something like the fact that the field is algebraically closed of characteristic prime to $|G|$)
