Tensor product of C* algebras

Question

Let $Y =[0,1]$. Consider the $C^*$-algebra $C(Y)$ acting on $L^{2}(Y).$ How to see $C(Y)\otimes C(Y)=C(Y\times Y)$?

Answer 1

It is not hard to check that $L^2(X)\otimes L^2(Y)\simeq L^2(X\times Y)$ as Hilbert spaces. A natural could be to map elementary tensors $f\otimes g\in L^2(X)\otimes L^2(Y)$ to the function $f\times g\in L^2(X\times Y)$ given by $(f\times g)(x,y)=f(x)g(y)$. It is easy to check that this map is linear, isometric, and onto, so it is a unitary $U$.

Now, as $C(X)\otimes C(Y)$ is the norm closure of the span of elementary tensors $f\otimes g$, we may define a map $\gamma:C(X)\odot C(Y)\to C(X\times Y)$ by mapping $f\otimes g\longmapsto f\times g$. Since we are looking at the continuous function acting as multiplication operators on $L^2$, we have \begin{align} \|\gamma(\sum_jf_j\otimes g_j)\|^2 &=\|\sum_j f_j\times g_j\|^2 =\sup\{\| (\sum_j f_j\times g_j)(\sum_s\eta_s\times \nu_s)\|^2:\ \|\sum_s\eta_s\times \nu_s\|=1\}\\ \ \\ &=\sup\{\sum_{j,k}\sum_{s,t}\int_{X\times Y} f_j\eta_s\,\overline{f_k\eta_t}\, g_j\nu_s\,\overline{g_k\eta_t}:\ \|\sum_s\eta_s\times \nu_s\|=1\}\\ \ \\ &=\sup\{\sum_{j,k}\sum_{s,t}\langle f_j\eta_s,f_k\eta_t\rangle\langle g_j\nu_s,g_k\eta_t\rangle:\ \|\sum_s\eta_s\times \nu_s\|=1\}\\ \ \\ &=\sup\{\sum_{j,k}\sum_{s,t}\langle (f_j\otimes g_j)(\eta_s\otimes \nu_s),(f_k\otimes g_k)(\eta_t\otimes \nu_t)\rangle:\ \|\sum_s\eta_s\otimes \nu_s\|=1\}\\ \ \\ &=\sup\{\|\sum_{j}\langle (f_j\otimes g_j)\sum_s \eta_s\otimes \nu_s\|^2:\ \|\sum_s\eta_s\otimes \nu_s\|=1\}\\ \ \\ &=\|\sum_j f_j\otimes g_j\|^2. \end{align} So $\gamma$ is an isometry. It clear that the image of $\gamma$ is dense, as functions of the form $\sum_j f_j\times g_j$ are dense in $C(X\times Y)$. Being an isometry, this makes $\gamma$ onto.