= \text{sot-}\lim_i \Phi(k_i).$$ Therefore we obtain $\Phi(x) = Q \Phi(x) P$ for all $x \in X$. We write $$\pi = P \Phi|_A P \, , \,\, t = Q \Phi|_X P \, , \,\, \psi = Q \Phi|_{{{\mathcal{K}}}(X)} Q.$$ Now $P$ and $Q$ are complementary projections and thus we get that $$\Phi(\begin{bmatrix} a & y^* \\ x & k \end{bmatrix}) = \begin{bmatrix} \pi(a) & t(y)^* \\ t(x) & \psi(k) \end{bmatrix}.$$ It is clear that $\pi$ and $\psi$ are \*-homomorphisms such that $\psi(k) t(x) \pi(a) = t(k(x)a)$ and that both $\pi$ and $t$ are complete isometries as $A, X \subseteq {{\mathcal{A}}}(X)$. We want to show that $\ker\Phi = (0)$. To reach contradiction let $$0 \neq f = \begin{bmatrix} a & y^* \\ x & k \end{bmatrix} \in \ker\Phi.$$ By applying $\Phi$ on $f^*f$ and restricting to the $(1,1)$-entry we get that $\pi(a^*a + {\left\langlex,x\right\rangle}) = 0$. But $\pi$ is a complete isometry as $A \subseteq {{\mathcal{A}}}(X)$ and thus $a=0$ and $x = 0$. Similarly applying to $f f^*$ gives that $y = 0$. By applying on $f g$ for any $g \in {{\mathfrak{L}}}(X)$ we also get that $t(k(z)) = 0$ for all $z \in X$. Hence we have that $k(z) = 