Maximal norm of tensor product Definition 3.3.3. (Maximal Norm) Given $A$ and B (two C*-algebra), we define the maximal C*-norm on $A \odot B$ to be 
$$||x||_{max}=sup\{||\pi(x)||: \pi: A\odot B\rightarrow B(H) ~a~*-homomorphism\}$$
for $x\in A\odot B$. We let $A\otimes_{max} B$ denote the completion of $A\odot B$ with respect to $||.||_{max}$. (Here, the "$\odot$" means algebraic tensor product.)
Proposition 3.3.7 (Universality). If $\pi: A \odot B \rightarrow C$ is a $*$-homomorphism, then there exists a unique $*$-homomorphism $A \otimes_{max} B \rightarrow C$ which extends $\pi$.
Proof. Faithfully representing $C$ on some Hilbert space, this fact follows from the definition of $||.||_{max}$.
However, I can not understand the proof of the Proposition. Yes, we have a $*$-homomorphism $A\odot B\rightarrow B(H)$ for some $H$. But how can we extend $\pi$ from the definition?

The proof looks too complicated.
Let $\sigma : C \hookrightarrow B(H)$ be a faithful $*$-representation of $C$. In particular $\sigma$ is isometric.
Then $\lVert \pi(x) \rVert = \lVert \sigma(\pi(x)) \rVert \leq \lVert x \rVert_{\mathrm{max}}$, since $\sigma \circ \pi$ is a $*$-rep. of $A \odot B$.
