How does one prove $C([0,1)\otimes A \cong C([0,1],A)$?

$$C([0,1])$$ is a $$C^*$$ algebra of complex functionals. $$A$$ is a $$C^*$$ algebra. Hence $$C([0,1],A)$$, the continuous functions from $$[0,1]$$ to $$A$$ is also a $$C^*$$ algebra.

We construct its tensor product in the category of $$C^*$$ algebras. This is unique as $$C([0,1])$$ is nuclear.

There is a canonical $$*$$-homomorphism, $$C[0,1] \otimes A \rightarrow C([0,1],A)$$ $$(f \otimes a) (x) = f(x)a$$ How does one show this is in fact an isomorphism?

The map is $$\pi:f\otimes a \longmapsto f a$$. It's obvious that it is linear and multiplicative, and preserves adjoints, so it is a $$*$$-homomorphism. It is injective: if $$\pi(\sum_j f_j\otimes a_j)=0$$, we may choose the $$a_j$$ so that they are linearlity independent. Then $$0=\pi(\sum_j f_j\otimes a_j)=\sum_j f_j a_j.$$ Evaluating at any $$t\in[0,1]$$, we have $$0=\sum_j f_j(t)a_j$$, and the linear independence gives $$f_j(t)=0$$ for all $$j$$. As we can do this for any $$t$$, $$f_j=0$$ for all $$j$$, and so $$\sum_jf_j\otimes a_j=0$$. Thus $$\pi$$ is injective (in reality one also needs to check that the map is well-defined, see this answer for details).
It remains to show that $$\pi$$ is onto. Since it is an isometry, it is enough to show that its range is dense. There you use compactness of $$[0,1]$$ to show that the functions of the form $$\sum_j f_j a_j$$ are dense. Concretely, let $$f\in C([0,1],A)$$, Fix $$\varepsilon>0$$, then there exists a partition $$0=t_0 such that $$\|f(t_j)-f(t)\|<\varepsilon$$ for $$t\in[t_j,t_{j+1}]$$. Let $$a_j=f(t_j)$$ for each $$j$$; let $$g_j=1_{[t_j,t_{j+1}]}$$, and let $$f_j\in C[0,1]$$ such that $$\|f_j-g_j\|<\varepsilon/(m\|f\|)$$. Then, with $$k$$ such that $$t\in[t_k,t_{k+1}]$$, \begin{align} \|f(t)-\sum_j f_j(t)a_j\| &\leq \|f(t)-\sum_j g_j(t)a_j\|+\|\sum_j(g_j(t)-f_j(t))a_j\|\\ &\leq \|f(t)-\sum_j g_j(t)a_j\|+\varepsilon\\ &=\|f(t)-f(t_k)\|+\varepsilon\\ \ \\ &<2\varepsilon. \end{align}
• Thanks a lot Martin? I wonder if you can comment on my recent $K$ theory question? – CL. Apr 13 at 21:45