[proofplan] We first construct the algebraic map from the bilinear multiplication rule and verify that it respects multiplication, involution, and the unit. We then identify the minimal tensor norm on $C(X)\odot C(Y)$ by using the faithful point-evaluation representations on $\ell^2(X)$ and $\ell^2(Y)$; under the spatial product representation, an algebraic tensor acts as multiplication by the corresponding function on $X\times Y$, so its operator norm is exactly the supremum norm. Finally, Stone-Weierstrass gives uniform density of the algebraic range in $C(X\times Y)$, and the isometry extends the algebraic map to a surjective and injective $*$-homomorphism on the minimal completion. [/proofplan] [step:Construct the algebraic tensor map from the pointwise product rule] Define a bilinear map \begin{align*} B:C(X)\times C(Y)\to C(X\times Y) \end{align*} by declaring that, for $f\in C(X)$, $g\in C(Y)$, and $(x,y)\in X\times Y$, \begin{align*} B(f,g)(x,y)=f(x)g(y). \end{align*} The function $B(f,g)$ is continuous because it is the pointwise product of the continuous functions $(x,y)\mapsto f(x)$ and $(x,y)\mapsto g(y)$ on $X\times Y$. By the universal property of the algebraic [tensor product](/page/Tensor%20Product), there is a unique [linear map](/page/Linear%20Map) \begin{align*} \Phi_0:C(X)\odot C(Y)\to C(X\times Y) \end{align*} such that $\Phi_0(f\otimes g)=B(f,g)$ for every $f\in C(X)$ and $g\in C(Y)$. Let $u\in C(X)\odot C(Y)$ and $v\in C(X)\odot C(Y)$ be finite sums \begin{align*} u=\sum_{i=1}^{m} f_i\otimes g_i \end{align*} and \begin{align*} v=\sum_{j=1}^{n} p_j\otimes q_j, \end{align*} where $m,n\in\mathbb N$, $f_i,p_j\in C(X)$, and $g_i,q_j\in C(Y)$. For every $(x,y)\in X\times Y$, \begin{align*} \Phi_0(uv)(x,y)=\sum_{i=1}^{m}\sum_{j=1}^{n} f_i(x)p_j(x)g_i(y)q_j(y). \end{align*} The right-hand side is exactly $\Phi_0(u)(x,y)\Phi_0(v)(x,y)$, so $\Phi_0(uv)=\Phi_0(u)\Phi_0(v)$. Also, \begin{align*} \Phi_0(u^*)(x,y)=\sum_{i=1}^{m}\overline{f_i(x)}\overline{g_i(y)}. \end{align*} This equals $\overline{\Phi_0(u)(x,y)}=\Phi_0(u)^*(x,y)$, so $\Phi_0(u^*)=\Phi_0(u)^*$. Since $\Phi_0(1_{C(X)}\otimes 1_{C(Y)})=1_{C(X\times Y)}$, the map $\Phi_0$ is a unital $*$-homomorphism. [/step] [step:Compute the minimal tensor norm by point-evaluation representations] Let $H_X:=\ell^2(X)$ and $H_Y:=\ell^2(Y)$, with standard orthonormal families $(e_x)_{x\in X}$ and $(e_y)_{y\in Y}$. Let $\mathcal{L}(H_X)$ and $\mathcal{L}(H_Y)$ denote the $C^*$-algebras of bounded linear operators on $H_X$ and $H_Y$, respectively, with the operator norm and Hilbert-space adjoint. Define representations \begin{align*} \pi_X:C(X)\to \mathcal{L}(H_X) \end{align*} and \begin{align*} \pi_Y:C(Y)\to \mathcal{L}(H_Y) \end{align*} by \begin{align*} \pi_X(f)e_x=f(x)e_x \end{align*} and \begin{align*} \pi_Y(g)e_y=g(y)e_y. \end{align*} These are unital $*$-representations. They are faithful because, if $f\in C(X)$ is nonzero, then there exists $x\in X$ with $f(x)\ne 0$, and hence $\pi_X(f)e_x=f(x)e_x\ne 0$; the same argument applies to $\pi_Y$. By [citetheorem:8580], the spatial product representation \begin{align*} \pi_X\odot\pi_Y:C(X)\odot C(Y)\to \mathcal{L}(H_X\otimes H_Y) \end{align*} is a well-defined $*$-homomorphism. By [citetheorem:8581], because $\pi_X$ and $\pi_Y$ are faithful, the minimal tensor norm is \begin{align*} \|u\|_{\min}=\|(\pi_X\odot\pi_Y)(u)\|_{\mathcal{L}(H_X\otimes H_Y)} \end{align*} for every $u\in C(X)\odot C(Y)$. Identify $H_X\otimes H_Y$ with $\ell^2(X\times Y)$ by the unitary map sending $e_x\otimes e_y$ to $e_{(x,y)}$. If \begin{align*} u=\sum_{i=1}^{m} f_i\otimes g_i, \end{align*} then, for every $(x,y)\in X\times Y$, \begin{align*} (\pi_X\odot\pi_Y)(u)e_{(x,y)}=\left(\sum_{i=1}^{m}f_i(x)g_i(y)\right)e_{(x,y)}. \end{align*} Thus $(\pi_X\odot\pi_Y)(u)$ is the diagonal multiplication operator on $\ell^2(X\times Y)$ by the function $\Phi_0(u)$. Since $X\times Y$ is compact and $\Phi_0(u)$ is continuous, $|\Phi_0(u)|$ attains its maximum. The operator norm of a diagonal multiplication operator on $\ell^2(X\times Y)$ is the supremum of the absolute values of its diagonal entries, so \begin{align*} \|(\pi_X\odot\pi_Y)(u)\|_{\mathcal{L}(H_X\otimes H_Y)}=\|\Phi_0(u)\|_{\infty}. \end{align*} Consequently, \begin{align*} \|u\|_{\min}=\|\Phi_0(u)\|_{\infty}. \end{align*} [guided] The main point is to prove that the norm used to complete $C(X)\odot C(Y)$ is exactly the [uniform norm](/page/Uniform%20Norm) of the corresponding function on $X\times Y$. We use the most concrete faithful representations available: point evaluations on Hilbert spaces with one basis vector for each point. Let $H_X:=\ell^2(X)$ and $H_Y:=\ell^2(Y)$. For each $x\in X$, let $e_x\in H_X$ denote the standard basis vector supported at $x$, and for each $y\in Y$, let $e_y\in H_Y$ denote the standard basis vector supported at $y$. Let $\mathcal{L}(H_X)$ and $\mathcal{L}(H_Y)$ denote the $C^*$-algebras of bounded linear operators on $H_X$ and $H_Y$, respectively, with the operator norm and Hilbert-space adjoint. Define \begin{align*} \pi_X:C(X)\to \mathcal{L}(H_X) \end{align*} by requiring \begin{align*} \pi_X(f)e_x=f(x)e_x \end{align*} for every $f\in C(X)$ and every $x\in X$. Similarly, define \begin{align*} \pi_Y:C(Y)\to \mathcal{L}(H_Y) \end{align*} by \begin{align*} \pi_Y(g)e_y=g(y)e_y. \end{align*} These are unital $*$-representations because products, sums, scalar multiples, units, and complex conjugates are all computed pointwise on the diagonal entries. They are faithful: if $f\ne 0$ in $C(X)$, then there exists $x\in X$ such that $f(x)\ne 0$, and then $\pi_X(f)e_x=f(x)e_x\ne 0$. Hence $\pi_X(f)\ne 0$. If $g\ne 0$ in $C(Y)$, then there exists $y\in Y$ such that $g(y)\ne 0$, and then $\pi_Y(g)e_y=g(y)e_y\ne 0$, so $\pi_Y(g)\ne 0$. Now [citetheorem:8580] applies to the two $*$-representations $\pi_X$ and $\pi_Y$, so the spatial product \begin{align*} \pi_X\odot\pi_Y:C(X)\odot C(Y)\to \mathcal{L}(H_X\otimes H_Y) \end{align*} is a well-defined $*$-homomorphism. Since both representations are faithful, [citetheorem:8581] says that the minimal tensor norm is computed from this representation: \begin{align*} \|u\|_{\min}=\|(\pi_X\odot\pi_Y)(u)\|_{\mathcal{L}(H_X\otimes H_Y)}. \end{align*} We next identify the operator explicitly. The Hilbert tensor product $H_X\otimes H_Y$ is naturally identified with $\ell^2(X\times Y)$ by the unitary map sending $e_x\otimes e_y$ to $e_{(x,y)}$. Let \begin{align*} u=\sum_{i=1}^{m} f_i\otimes g_i \end{align*} be an arbitrary element of $C(X)\odot C(Y)$, where $m\in\mathbb N$, $f_i\in C(X)$, and $g_i\in C(Y)$. Then, for every $(x,y)\in X\times Y$, \begin{align*} (\pi_X\odot\pi_Y)(u)e_{(x,y)}=\left(\sum_{i=1}^{m}f_i(x)g_i(y)\right)e_{(x,y)}. \end{align*} The scalar multiplying $e_{(x,y)}$ is exactly $\Phi_0(u)(x,y)$. Thus $(\pi_X\odot\pi_Y)(u)$ is the diagonal multiplication operator by $\Phi_0(u)$. It remains to compute the norm of this diagonal operator. Since $X\times Y$ is compact and $\Phi_0(u)$ is continuous, there is a point $(x_0,y_0)\in X\times Y$ such that \begin{align*} |\Phi_0(u)(x_0,y_0)|=\|\Phi_0(u)\|_{\infty}. \end{align*} For every vector $\xi\in \ell^2(X\times Y)$, the diagonal form gives \begin{align*} \|(\pi_X\odot\pi_Y)(u)\xi\|_{\ell^2(X\times Y)}\le \|\Phi_0(u)\|_{\infty}\|\xi\|_{\ell^2(X\times Y)}. \end{align*} Hence the operator norm is at most $\|\Phi_0(u)\|_{\infty}$. Applying the operator to the unit vector $e_{(x_0,y_0)}$ gives \begin{align*} \|(\pi_X\odot\pi_Y)(u)e_{(x_0,y_0)}\|_{\ell^2(X\times Y)}=\|\Phi_0(u)\|_{\infty}. \end{align*} Therefore the operator norm is also at least $\|\Phi_0(u)\|_{\infty}$. Combining the two inequalities gives \begin{align*} \|(\pi_X\odot\pi_Y)(u)\|_{\mathcal{L}(H_X\otimes H_Y)}=\|\Phi_0(u)\|_{\infty}. \end{align*} Since this operator norm is the minimal tensor norm, we obtain \begin{align*} \|u\|_{\min}=\|\Phi_0(u)\|_{\infty}. \end{align*} [/guided] [/step] [step:Show the algebraic range is uniformly dense in $C(X\times Y)$] Let $\mathcal{A}:=\Phi_0(C(X)\odot C(Y))\subset C(X\times Y)$. Since $\Phi_0$ is a unital $*$-homomorphism, $\mathcal{A}$ is a unital self-adjoint subalgebra of $C(X\times Y)$. We show that $\mathcal{A}$ separates points. Let $(x_1,y_1),(x_2,y_2)\in X\times Y$ be distinct. If $x_1\ne x_2$, then by the Urysohn separation lemma for compact Hausdorff spaces, there exists $f\in C(X)$ such that $f(x_1)\ne f(x_2)$. Taking $g:=1_{C(Y)}$, we get \begin{align*} \Phi_0(f\otimes g)(x_1,y_1)=f(x_1) \end{align*} and \begin{align*} \Phi_0(f\otimes g)(x_2,y_2)=f(x_2), \end{align*} so $\mathcal{A}$ separates these two points. If $x_1=x_2$, then necessarily $y_1\ne y_2$. By the Urysohn separation lemma for compact Hausdorff spaces, there exists $g\in C(Y)$ such that $g(y_1)\ne g(y_2)$. Taking $f:=1_{C(X)}$, the function $\Phi_0(f\otimes g)$ separates $(x_1,y_1)$ and $(x_2,y_2)$. Thus $\mathcal{A}$ is a unital self-adjoint subalgebra of $C(X\times Y)$ that separates points. By the complex [Stone-Weierstrass theorem](/theorems/886) for compact Hausdorff spaces, $\mathcal{A}$ is dense in $C(X\times Y)$ with respect to the supremum norm. [/step] [step:Extend the algebraic map to an isometric surjective $*$-isomorphism] The identity \begin{align*} \|u\|_{\min}=\|\Phi_0(u)\|_{\infty} \end{align*} for every $u\in C(X)\odot C(Y)$ shows that $\Phi_0$ is an isometry from the normed $*$-algebra $C(X)\odot C(Y)$ with the minimal tensor norm into $C(X\times Y)$. Therefore $\Phi_0$ extends uniquely by continuity to an isometric linear map \begin{align*} \Phi:C(X)\otimes_{\min}C(Y)\to C(X\times Y). \end{align*} Because multiplication and involution are continuous in both $C^*$-norms and $\Phi_0$ is a unital $*$-homomorphism on the dense algebraic tensor product, the extension $\Phi$ is also a unital $*$-homomorphism. The range of $\Phi$ is closed because $\Phi$ is an isometry from a complete normed space into $C(X\times Y)$. It contains $\Phi_0(C(X)\odot C(Y))$, which is dense in $C(X\times Y)$ by the preceding step. Hence the range of $\Phi$ is all of $C(X\times Y)$. Finally, since $\Phi$ is an isometry, $\ker \Phi=\{0\}$. Thus $\Phi$ is a bijective unital $*$-homomorphism, and therefore an isometric $*$-isomorphism \begin{align*} C(X)\otimes_{\min}C(Y)\cong C(X\times Y). \end{align*} [/step]