A Lemma about the operator space

The following lemma comes from the book "C*-algebras Finite-Dimensional Approximations" by N.P. Brown and N. Ozawa P379

Lemma 13.2.3 Let $X_{i}\in B(H_{i})$ (i=1,2) be unital operator subspaces and let $\phi: X_{1}\rightarrow X_{2}$ be a unital complete isomery. Suppose that $X_{2}$ is spanned by unitary elements in $B(H_{2})$. Then, $\phi$ uniquely extends to a -homomorphism between the C-subalgebras $C^{*}(X_{1})$ generated by $X_{i}$ in $B(H_{i})$.

Proof By Arveson's Extension Theorem, $\phi$ extends to a complete contractive map (i.e. $||\phi||_{cb}\leq1$) from $B(H_{1})\rightarrow B(H_{2})$, which we still denote by $\phi$. Since $\phi$ is unital, it has to be a u.c.p map. Since $\phi|_{X_{1}}$ is isometric and $X_{2}$ is spanned by unitary elements, $X_{1}$ is contained in the multiplicative domain of $\phi$. Hence, $\phi$ is a *-homomorphism on $C^{*}(X_{1})$.

In the proof of this lemma, the author first use the "operator space" version of Arveson's Extension Theorem to get a c.c map $\phi$ from $B(H_{1})\rightarrow B(H_{2})$. And then comes my question:

1. The author says "Since $\phi$ is unital, it has to be a u.c.p." Why?

2. The author says "Since $\phi|_{X_{1}}$ is isometric and $X_{2}$ is spanned by unitary elements, $X_{1}$ is contained in the multiplicative domain of $\phi$." Why?

1. If $x\geq0$, let $f$ be a state on $B(H_2)$; then the functional $f\circ\phi$ is unital and contractive. It is well-known that a unital contractive functional is positive (I know of proofs in Paulsen and Davidson's books). So $f(\phi(x))\geq0$ for all states on $B(H_2)$, which means that $\phi(x)\geq0$. Finally, this can be done for each amplification of $\phi$, and so $\phi$ is cp.
2. Let $u\in B(H_2)$ be a unitary, $w=\phi^{-1}(u)$. Then $\|w\|=1$, so $w^*w\leq I$ and $$I=u^*u=\phi(w)^*\phi(w)\leq\phi(w^*w)\leq\phi(I)=I.$$ Then $\phi(w)^*\phi(w)=\phi(w^*w)$, so $w$ is in the multiplicative domain of $\phi$. Since $\phi$ is in particular a vector space isomorphism between $X_1$ and $X_2$, these $w$ (the preimages of unitaries in $X_2$) span $X_1$, and so $\phi$ is multiplicative on $C^*(X_1)$.