[proofplan] We construct $\pi\odot\rho$ from the universal property of the algebraic [tensor product](/page/Tensor%20Product), using the bilinear map $(a,b)\mapsto \pi(a)\otimes\rho(b)$. The main point is then to check the algebraic operations on elementary tensors: multiplication follows from the product rule for tensor products of bounded operators, and the involution follows from the adjoint rule. Since every element of $A\odot B$ is a finite sum of elementary tensors, bilinearity and conjugate-linearity extend the identities to all of $A\odot B$. [/proofplan]

[step:Construct the linear map from the bilinear operator tensor map] For each $a\in A$ and $b\in B$, the operators $\pi(a)\in\mathcal{L}(H)$ and $\rho(b)\in\mathcal{L}(K)$ determine a bounded operator $\pi(a)\otimes\rho(b)\in\mathcal{L}(H\otimes K)$. Define \begin{align*} \Phi:A\times B\to \mathcal{L}(H\otimes K),\qquad (a,b)\mapsto \pi(a)\otimes\rho(b). \end{align*} The map $\Phi$ is complex bilinear because $\pi$ and $\rho$ are complex linear and the bounded-operator tensor product is bilinear in the two operator variables. By the universal property of the algebraic tensor product, there is a unique complex [linear map](/page/Linear%20Map) \begin{align*} \pi\odot\rho:A\odot B\to \mathcal{L}(H\otimes K) \end{align*} such that \begin{align*} (\pi\odot\rho)(a\otimes b)=\Phi(a,b)=\pi(a)\otimes\rho(b) \end{align*} for all $a\in A$ and $b\in B$. This proves well-definedness and uniqueness. [/step]

[step:Verify multiplicativity on elementary tensors]Let $a_1,a_2\in A$ and $b_1,b_2\in B$. For elementary Hilbert tensors $h\otimes k\in H\odot K$, the product rule for tensor products of bounded operators gives \begin{align*} ((\pi(a_1)\otimes\rho(b_1))(\pi(a_2)\otimes\rho(b_2)))(h\otimes k)=(\pi(a_1)\pi(a_2)h)\otimes(\rho(b_1)\rho(b_2)k). \end{align*} Since $\pi$ and $\rho$ are homomorphisms, \begin{align*} \pi(a_1)\pi(a_2)=\pi(a_1a_2) \end{align*} and \begin{align*} \rho(b_1)\rho(b_2)=\rho(b_1b_2). \end{align*} Thus the two bounded operators \begin{align*} (\pi(a_1)\otimes\rho(b_1))(\pi(a_2)\otimes\rho(b_2)) \end{align*} and \begin{align*} \pi(a_1a_2)\otimes\rho(b_1b_2) \end{align*} agree on the dense subspace $H\odot K\subset H\otimes K$, so they are equal. Therefore \begin{align*} (\pi\odot\rho)((a_1\otimes b_1)(a_2\otimes b_2))=(\pi\odot\rho)(a_1\otimes b_1)(\pi\odot\rho)(a_2\otimes b_2). \end{align*}[/step]

[guided]We first check multiplication in the simplest possible case, namely on elementary tensors. This is enough because the multiplication on $A\odot B$ is defined by bilinear extension from elementary products, and every element of $A\odot B$ is a finite sum of such tensors. Take $a_1,a_2\in A$, $b_1,b_2\in B$, $h\in H$, and $k\in K$. The elementary tensor $h\otimes k$ belongs to the algebraic tensor product $H\odot K$, which is dense in the [Hilbert space](/page/Hilbert%20Space) tensor product $H\otimes K$. By the defining action of tensor products of bounded operators, \begin{align*} ((\pi(a_1)\otimes\rho(b_1))(\pi(a_2)\otimes\rho(b_2)))(h\otimes k)=(\pi(a_1)\pi(a_2)h)\otimes(\rho(b_1)\rho(b_2)k). \end{align*} Because $\pi:A\to\mathcal{L}(H)$ and $\rho:B\to\mathcal{L}(K)$ are representations, they preserve multiplication. Hence \begin{align*} \pi(a_1)\pi(a_2)=\pi(a_1a_2) \end{align*} and \begin{align*} \rho(b_1)\rho(b_2)=\rho(b_1b_2). \end{align*} Substituting these identities into the previous formula gives \begin{align*} ((\pi(a_1)\otimes\rho(b_1))(\pi(a_2)\otimes\rho(b_2)))(h\otimes k)=(\pi(a_1a_2)h)\otimes(\rho(b_1b_2)k). \end{align*} The right-hand side is exactly \begin{align*} (\pi(a_1a_2)\otimes\rho(b_1b_2))(h\otimes k). \end{align*} Thus the two bounded operators agree on $H\odot K$. Since $H\odot K$ is dense in $H\otimes K$ and bounded operators agreeing on a dense subspace agree everywhere, we obtain \begin{align*} (\pi(a_1)\otimes\rho(b_1))(\pi(a_2)\otimes\rho(b_2))=\pi(a_1a_2)\otimes\rho(b_1b_2). \end{align*} Using the defining formula for $\pi\odot\rho$ and the product rule \begin{align*} (a_1\otimes b_1)(a_2\otimes b_2)=a_1a_2\otimes b_1b_2 \end{align*} in $A\odot B$, this becomes \begin{align*} (\pi\odot\rho)((a_1\otimes b_1)(a_2\otimes b_2))=(\pi\odot\rho)(a_1\otimes b_1)(\pi\odot\rho)(a_2\otimes b_2). \end{align*}[/guided]

[step:Extend multiplicativity from elementary tensors to finite sums] Let $x,y\in A\odot B$. Choose finite representations \begin{align*} x=\sum_{i=1}^{m}a_i\otimes b_i \end{align*} and \begin{align*} y=\sum_{j=1}^{n}c_j\otimes d_j \end{align*} with $m,n\in\mathbb N$, $a_i,c_j\in A$, and $b_i,d_j\in B$. By bilinearity of multiplication in $A\odot B$, \begin{align*} xy=\sum_{i=1}^{m}\sum_{j=1}^{n}(a_ic_j)\otimes(b_id_j). \end{align*} Using linearity of $\pi\odot\rho$ and the elementary tensor computation from the previous step, \begin{align*} (\pi\odot\rho)(xy)=\sum_{i=1}^{m}\sum_{j=1}^{n}(\pi(a_i)\pi(c_j))\otimes(\rho(b_i)\rho(d_j)). \end{align*} By distributivity of operator multiplication in $\mathcal{L}(H\otimes K)$, this equals \begin{align*} \left(\sum_{i=1}^{m}\pi(a_i)\otimes\rho(b_i)\right)\left(\sum_{j=1}^{n}\pi(c_j)\otimes\rho(d_j)\right). \end{align*} Therefore \begin{align*} (\pi\odot\rho)(xy)=(\pi\odot\rho)(x)(\pi\odot\rho)(y). \end{align*} [/step]

[step:Verify preservation of the involution] Let $a\in A$ and $b\in B$. For $h,h'\in H$ and $k,k'\in K$, the Hilbert tensor product [inner product](/page/Inner%20Product) satisfies \begin{align*} ((\pi(a)\otimes\rho(b))(h\otimes k),h'\otimes k')_{H\otimes K}=(\pi(a)h,h')_H(\rho(b)k,k')_K. \end{align*} Using the adjoint identities for $\pi(a)$ and $\rho(b)$, this becomes \begin{align*} (h,\pi(a)^*h')_H(k,\rho(b)^*k')_K. \end{align*} Since $\pi$ and $\rho$ are $*$-representations, \begin{align*} \pi(a)^*=\pi(a^*) \end{align*} and \begin{align*} \rho(b)^*=\rho(b^*). \end{align*} The preceding inner product identity first holds on elementary tensors and then, by finite linearity in each vector variable, on all pairs of vectors in $H\odot K$. Since $H\odot K$ is dense in $H\otimes K$ and both candidate adjoints are bounded operators, equality of the corresponding inner products on $H\odot K$ extends to all of $H\otimes K$. Hence \begin{align*} (\pi(a)\otimes\rho(b))^*=\pi(a^*)\otimes\rho(b^*). \end{align*} Using the involution on $A\odot B$, this gives \begin{align*} (\pi\odot\rho)((a\otimes b)^*)=(\pi\odot\rho)(a\otimes b)^*. \end{align*} Now let $x\in A\odot B$ and choose a finite representation \begin{align*} x=\sum_{i=1}^{m}\lambda_i(a_i\otimes b_i) \end{align*} with $m\in\mathbb N$, $\lambda_i\in\mathbb C$, $a_i\in A$, and $b_i\in B$. The involution on $A\odot B$ is conjugate-linear, so \begin{align*} x^*=\sum_{i=1}^{m}\overline{\lambda_i}(a_i\otimes b_i)^*. \end{align*} Using linearity of $\pi\odot\rho$, the elementary tensor identity, and conjugate-linearity of the adjoint on $\mathcal{L}(H\otimes K)$, we obtain \begin{align*} (\pi\odot\rho)(x^*)=\sum_{i=1}^{m}\overline{\lambda_i}(\pi\odot\rho)((a_i\otimes b_i)^*) \end{align*} and therefore \begin{align*} (\pi\odot\rho)(x^*)=\sum_{i=1}^{m}\overline{\lambda_i}(\pi\odot\rho)(a_i\otimes b_i)^*. \end{align*} By conjugate-linearity of the adjoint again, \begin{align*} \sum_{i=1}^{m}\overline{\lambda_i}(\pi\odot\rho)(a_i\otimes b_i)^*=\left(\sum_{i=1}^{m}\lambda_i(\pi\odot\rho)(a_i\otimes b_i)\right)^*. \end{align*} Thus \begin{align*} (\pi\odot\rho)(x^*)=(\pi\odot\rho)(x)^*. \end{align*} Therefore $\pi\odot\rho$ preserves the involution. [/step]

[step:Conclude that the spatial product is a star homomorphism] The map $\pi\odot\rho$ is linear by construction. The preceding steps show that it preserves multiplication and the involution on all of $A\odot B$. Therefore \begin{align*} \pi\odot\rho:A\odot B\to\mathcal{L}(H\otimes K) \end{align*} is a well-defined $*$-homomorphism satisfying \begin{align*} (\pi\odot\rho)(a\otimes b)=\pi(a)\otimes\rho(b) \end{align*} for every $a\in A$ and $b\in B$. [/step]