Let $A=C([0,1])$ be an abstract $C^{*}$-algebra. Let $(t_i)_{i=1}^{\infty}$ be a dense sequence in $[0,1]$, using this, we define $\phi:A\to\mathcal{B}(\ell^2(\mathbb{N}))$ as below: $$\phi(a)=\begin{bmatrix} a(t_1) & 0 & 0 & \dots \\ 0 & a(t_2) & 0 &\dots \\ 0 & 0 & a(t_3) &\dots \\ \vdots & \vdots & \ddots \end{bmatrix}$$
Then we define $\psi:A\to\mathcal{B}(L^2([0,1]))$ as $\Big(\psi(a)(f)\Big)(t)=a(t)f(t)$. I already proved both $\phi$ and $\psi$ are injective $*$-homomorphisms, now I am asked to show that there doesn't exist a unitary operator $u:L^2([0,1])\to\ell^2(\mathbb{N})$ such that $u^{*}\phi(a)u=\psi(a)$ for all $a\in A$.
I have no idea how to do that, any help please......
If every unitary operator is an surjection (or isomorphism or bijection), then it remains to apply the reasoning below.
Let $a_0(t)=t-t_1$. Then $\phi (a_0)(e_1)=a_0(t_1)e_1=0$, where $e_1=(1,0,0,...)$. Сhoose $f\in L^2[0,1]$ such that $u(f)=e_1$ ($f$ exists because $u$ is an surjection). It follows from $$u^*\phi(a_0)u(f)=u^*\phi(a_0)(e_1)=u^*(0)=0$$ that $\operatorname{Ker}\big(u^*\phi(a_0)u\big)\neq \{0\}$.
But if $\psi(a_0)(f)=0$ then $(t-t_1)f(t)=0$ for a. e. $t$. Since $t-t_1=0$ at one point , $f=0$ a. e. So $\operatorname{Ker}\big(\psi(a_0)\big)=\{0\}\neq\operatorname{Ker}\big(u^*\phi(a_0)u\big)$. Therefore $u^*\phi(a_0)u\neq \psi(a_0)$.