The tensor product of lie group representative functions is equal to the representative functions of the product group I would like to prove the following lemma:
Let $\mathscr{T}(G,\mathbb{C})$ be the $\mathbb{C}$-algebra of representative functions from a lie group $G$ to $\mathbb{C}$. Then the $\mathbb{C}$-algebra homomorphism $$t:\mathscr{T}(G,\mathbb{C})\otimes_\mathbb{C}\mathscr{T}(H,\mathbb{C})\rightarrow \mathscr{T}(G\times H,\mathbb{C})$$ sending $u\otimes v$ to $(g,h)\mapsto u(g)v(h)$ is an isomorphism.
It seems to me the way to do this is to show that any representative function $f\in \mathscr{T}(G\times H,\mathbb{C})$ can be written as a linear combination of product functions $(g,h)\mapsto u(g)v(h)$. I have been trying to use that face that $f$ generates a finite dimensional subspace of $C^0(G,\mathbb{C})$, but I'm not sure how to use that to characterize $f$ itself. Honestly, I have no idea where to start, and any help would be appreciated.
 A: I'm a bit rusty on the details of this argument, but I know that it goes something along the following lines.

Suppose that $f \in \mathscr{T}(G \times H, \mathbb{C})$. Then the $\mathbb{C}(G \times H)$-module, $f \circ C^0(G \times H, \mathbb{C})$, given by 
$$((g,h) \cdot f)(x,y) = f(xg,yh),
$$ 
for all $g,h,x,y \in G$, and $f \in C^0(G \times H, \mathbb{C})$ is finite dimensional. So there exists some $\mathbb{C}$-basis 
$$
f_1, \dots, f_n \in C^0(G \times H, \mathbb{C}),
$$ 
and without loss of generality we may take $f = f_1$. Now each 
$$
f_i \circ C^{0}(G \times H, \mathbb{C}) \subseteq f \circ C^0(G \times H, \mathbb{C}),
$$
and so each $f_i$ is finitary. Then
$$
(g,h) \cdot f_j = \sum_{i=1}^{n} c_{i,j}(g,h) f_i,
$$
for some $c_{i,j}(g,h) \in \mathbb{C}$. It can be seen that these $c_{i,j} : G \times H \to \mathbb{C}$ are in $C^0(G \times H, \mathbb{C})$. Moreover, by associativity we must have that
$$
((g',h') \cdot c_{i,k})((g,h)) =  c_{i,k}((g,h)(g',h')) = \sum_{j} c_{i,j}((g,h))c_{j,k}((g',h')),
$$
and so
$$
 (g',h') \cdot c_{i,k} = \sum_j c_{j,k}((g',h')) c_{i,j},
$$
and so $\left\{c_{i,j}\right\}_j$ is a $\mathbb{C}$-spanning set for 
$$
c_{i,k} \circ C^0(G \times H, \mathbb{C}),
$$
and so the $c_{i,k}$ are also finitary. Now the maps
\begin{align*}
 G_i  : & G \to \mathbb{C} & H_i : &  H \to \mathbb{C} \\
 & g \mapsto f_i(g,1) & &  h \mapsto   c_{i,1}(1,h),
\end{align*}
are finitary elements of $C^0(G, \mathbb{C})$, and $C^0(H, \mathbb{C})$ respectively. Then
\begin{align*}
t\left(\sum_i G_i \otimes H_i\right)(g,h) & = \sum_i t\left( G_i \otimes H_i \right)(g,h) \\
& = \sum_i G_i(g) H_i(h) \\
& = \sum_i f_i(g,1) c_{i,1}(1,h) \\
& = \left( \sum_i  c_{i,1}(1,h) f_i \right) (g,1)  \\
& = ((1,h) \cdot f_1)(g,1) = f_1(g,h) = f(g,h),
\end{align*}
and so $t$ is surjective. Can you show that it is injective?
