For every projection $p$ and normal $a$ in a C*-algebra $A$ (with $ap=pa$), there is a $*$-isomorphism $C(\sigma(a))\to C^{*}(a,p)$ such that ... Let $a$ be a normal element of a (non-unital) C*-algebra $A$. I am trying to prove that for every projection $p\in A$ that commutes with $a$ (i.e. $p=p^{2}=p^{*}$ and $ap=pa$), there is a $*$-isomorphism $\phi:C(\sigma(a))\to C^{*}(a,p)$ such that $\phi(z)=a$ and $\phi(1)=p$. 
Summary of the notation:


*

*$\sigma(a)$ is the spectrum of $A$,

*$z\colon\sigma(a)\to\mathbb{C}$ is the inclusion map,

*$C^{*}(a,p)$ is the C*-subalgebra of $A$ generated by $a$ and $p$.



My attempt: I think that $p$ is a unit for $C^{*}(a,p)$ (but I am not able to prove it yet). If $C^{*}(a)_{1}$ is the unitization of $C^{*}(a)$ (= the C*-subalgebra of $A$ generated by $a$), then the map 
  $$\pi_{1}\colon C^{*}(a)_{1}\to C^{*}(a,p),\qquad x+\lambda\mapsto x+\lambda p$$
  is a $*$-isomorphism. Note that $a$ is normal. Let $\pi_{2}\colon C(\sigma(a))\to C^{*}(a)_{1}$ be the functional calculus of $a$ in $C^{*}(a)_{1}$ (note that $\sigma(a)$ does not depend on the C*-subalgebra of $A$ containing $a$). Then $\pi_{1}\circ\pi_{2}(1)=p$ and $\pi_{1}\circ\pi_{2}(z)=a$. Hence $\phi:=\pi_{1}\circ\pi_{2}$ should do the job.

So (I think that) this proof only works if $p$ is unit for $C^{*}(a,p)$. I still haven't used that $ap=pa$, so this is probably needed to prove my claim (if my claim is true). Any suggestions are greatly appreciated!
 A: For your $\pi_1$ to be  multiplicative, you need $ap=a$. Otherwise, you could have for instance that $ap=p$; in that case, 
$$
\pi_1((a+\lambda)^2)=\pi_1(a^2+2\lambda a+\lambda^2)=a^2+2\lambda a+\lambda^2 p,
$$
while
$$
\pi_1(a+\lambda)^2=(a+\lambda p)^2=a^2+(2\lambda+\lambda^2)p. 
$$
And you have already used that $ap=pa$; if not, then you couldn't possibly get that $\pi_1$ is  multiplicative. 
When $pa=a$, then $C^*(a,p)$ is a unitization of $C^*(a)$, so the isomorphism is automatic. 
In general, it is not true that $C(\sigma(a))\simeq C^*(a,p)$. Let $A=\mathbb C^4$, and 
$$
a=(2,1,0,1),\ \ \ p=(0,1,1,0). 
$$
Then $pa=ap=(0,1,0,0)$. 
We have $\sigma(a)=\{0,1,2\}$, so $C(\sigma(a))=\mathbb C^3$. The algebra $C^*(a,p)$, on the other hand, contains $(1,0,0,0)=f(a)$ for $f$ a polynomial with $f(2)=1$, $f(1)=f(0)=0$. Also $(0,1,0,0)=pa$. Also $(0,0,1,0)=p-pa$. And $(0,0,0,1)=a-2(1,0,0,0)-(0,1,0,0)$. So $C^*(a,p)=\mathbb C^4$. 
If you want the above $A$ to be non-unital, just use $A=\mathbb C^4\oplus c_0$ and work on the first four coordinates. 
