Im trying to understand proof of corollary $11.34$ from here. The corollary goes as follows:
Let $A_1$ and $A_2$ be $C^{\ast}$-algebras. Given any $C^{\ast}$-norm $\vert \vert . \vert \vert$on $A_1 \otimes A_2$, there is a surjective ${\ast}$-homomorphism, $A_1 \otimes ^{\text{max}} A_2 \to \overline{A_1\otimes A_2}^{\vert \vert .\vert \vert}$ extending the identity map on $A_1 \otimes A_2$.
Can someone explain me why the identity map $A_1 \otimes A_2 \to \overline{A_1\otimes A_2}^{\vert \vert .\vert \vert}$ is a contraction to extend it to desired space?
I will use the notations from loc. cit. - and start with $A_1,A_2$ as in loc. cit. and a $C^*$-norm $\|\ \cdot\ \|$ on the algebraic tensor product $B_0:=A_1\odot A_2$. Let $C$ be the completion of $B_0$ w.r.t. this norm, so $(C, \|\ \cdot\ \|)$ is explicitly in that long notation $\left(\overline{A_1\odot A_2}^{\|\ \cdot\ \|}\ ,\ \|\ \cdot\ \|\right)$.
Now on $B_0:=A_1\odot A_2$ we also have the $\max$-norm $\|\ \cdot\ \|_\max$, and the completion w.r.t. this norm is $(B, \|\ \cdot\ \|_\max)=(A_1\otimes_\max A_2, \|\ \cdot\ \|_\max)$.
We define from $(B_0,\|\ \cdot\ \|_\max)$ a map $\pi$ to $(C,\|\ \cdot\ \|)$ by $\pi x = x$. Now we are in the situation of Proposition 11.33. The $*$-algebra $B_0$ is dense in $B$. The map $\pi$ is a $*$-homomorphism. And $\pi$ is a contraction from $(B_0, \|\ \cdot\ \|_\max)$ to $(C,\|\ \cdot\ \|)$ since for any $x\in B_0$ we have $$ \|\pi x\|_C=\|x\|_C=\|x\|\ {\color{blue}\le}\ \|x\|_\max=\|x\|_B\ . $$ The blue inequality comes from the claim in the sentence immediately preceding Cor. 11.34.