Isomorphism $C_0(X) \otimes A \to C_0(X,A)$ Let $A$ be a $C^*$-algebra. I want to prove that 
$$
\phi : C_0(X) \otimes_\alpha A \to C_0(X,A): f \otimes a \mapsto a \cdot f
$$
is an isomorphism, where $\alpha$ is a norm making the tensor product into a $C^*$-algebra. Note that $C_0(X)$ is commutative hence nuclear. Since $\phi(C_0(X) \otimes A)$ is dense (with $\otimes$ the algebraic tensor product) one is done, if the restriction of   $\phi $ to $C_0(X) \otimes A$ is an isometry. Then $\phi$ is clearly surjective and also injective. (On the algebraic tensorproduct $\phi$ is injective but I think this does not necessarily  extend to the completion.) 
So my question would be what one can take for $\alpha$ to prove in an efficient way that $\phi$ is an isometry.
 A: By construction, $\phi$ is a $*$-epimorphism, if it is well-defined. The argument for well-definedness, as usually happens, is tied with the proof that $\phi$ is one-to-one.
The key fact is the following:

Lemma. The following statements are equivalent:


*

*$\sum a_j\otimes f_j=0$




*There exists $T\in M_n(\mathbb C)$ such that $\sum_kT_{kj}a_k=0$, and $\sum_jT_{kj}f_j=f_k$.

Now, if $\sum a_j\otimes f_j=0$, we take $T$ from the Lemma, and then
$$
\sum_ka_kf_k=\sum_k\sum_jT_{kj}a_kf_j=\sum_j\left(\sum_kT_{kj}a_k\right)f_j=0.
$$
So $\phi$ is well-defined.
Conversely, if $\sum_ja_jf_j=0$, by re-arranging the sum we may assume that the $a_j$ are linearly independent. But then, for any any $t\in X$, $\sum f_j(t)a_j=0$, which implies $f_j(t)=0$ by the linear independence. Thus $f_j=0$ and so $\sum a_j\otimes f_j=0$ (in the re-arranged version, but this is implies the original version is also zero).
So $\phi$ is one-to-one. We can use this to define $$\|x\|_\alpha:=\|\phi(x)\|.$$ Since $\alpha$ is a norm and $C_0(X)$ is nuclear, we get that $\|x\|_\alpha=\|x\|$, i.e. that $\phi$ is isometric.
An isomorphism is isometric, so as expected $\|x\|_\alpha=\|\phi(x)\|$ for any $x\in C_0(X)\otimes_\alpha A$ and any tensor norm you could define on $C_0(X)\otimes A$.

To address the question of why $\phi$ is surjective: since it is a $*$-homomorphism it's image is closed, so it is enough to show density. Fix $f\in C_0(X,A)$ and $\varepsilon>0$. There exists $K$ compact such that $\|f\|<\varepsilon$ outside of $K$. For each $x$ there exists, by continuity, an open neighbourhood $V_x$ of $x$ such that $\|f(y)-f(x)\|<\varepsilon$ for all $y\in V_x$. By compactness we can cover $K$ with $V_{x_1},\ldots,V_{x_m}$.
Let $\{g_j\}$ be a partition of unity subordinated to $\{V_{x_j}\}$, and let $$g(t)=\sum_jf(x_j)\,g_j(t).$$ Being a sum of continuous functions, $g$ is continuous. For $t\in\bigcup_jV_{x_j},$ $$\|f(t)-g(t)\|=\Big\|\sum_j(f(t)-f(x))g_j(t)\Big\|\leq\varepsilon\sum_jg_j(t)=\varepsilon.$$ And for $t$ outside the open cover, $g(t)=0$ and $\|f(t)\|<\varepsilon$. So $\|f-g\|<\varepsilon$.
