If $\pi: A\rightarrow B(H)$ and $\sigma: B\rightarrow B(K)$ are arbitrary representations, prove that there exists a unique extending -homomorphism $\pi\otimes\sigma: A\otimes B\rightarrow B(H\otimes K)$ such that $\pi\otimes \sigma(a\otimes b)=\pi(a)\otimes\sigma(b)$. (The $A\otimes B$ is the spatial tensor product of C-algebra $A$ and $B$)
This is an exercise from the book "C*-algebras and Finite Dimensional Approximations". The hint is dilating $\pi$ and $\sigma$ to faithful repersentations and then cut back down. But how to dilate $\pi$ and $\sigma$ to faithful repersentations？
One can easily define $\pi\otimes\sigma$ on $A\odot B$, with the only slight difficulty in checking that it is well defined. The problem is that we need it to be bounded in order to extend to the completion.
If both representations are faithful then there is no issue with this, since $\rho_A\otimes\rho_B$ is isometric on $A\otimes B$ for any pair of faithful representations.
Fix faithful representations $\rho_A$ and $\rho_b$. Let $\tilde\pi:A\to B(H\oplus H_A)$, $\tilde\sigma:B\to B (H\oplus H_B)$ be given by $$ \tilde\pi(a)=\pi(a)\oplus \rho_A(a),\qquad\,\tilde\sigma(b)=\sigma(b)\oplus\rho_B(b). $$ Now $\tilde\pi\otimes\tilde\sigma$ is well-defined and isometric. Then, for any $x\in A\odot B$, \begin{align} \|(\pi\otimes\sigma)(x)\| &=\big\|\big((P_H\oplus0)\otimes(P_H\oplus0)\big)(\tilde\pi\otimes\tilde\sigma)(x)\,\big((P_H\oplus0)\otimes(P_H\oplus0)\big)\big\|\\[0.3cm] &\leq\|(\tilde\pi\otimes\tilde\sigma)(x)\|\leq\|x\|. \end{align}