Confusion over Continuous Functional Calculus I have the following problem:  As far as I understand, the continuous functional calculus gives an isomorphism from the continuous functions over the spectrum of a normal element to the $C^*$-algebra that the normal element and the unit generate. More specifically, if $A$ is a unital $C^*$-algebra and $a\in A$ is normal, then we have a star-isomorphism $\Gamma_A:C(\sigma_A(a))\cong C^*(a,1_A)$ so that $\Gamma_A(1)=1_A$ and $\Gamma_A(z)=a$, where $z$ is the identity map.
Suppose now that $B\subset A$ is another unital $C^*$-algebra (with unit $1_B$) that contains the above element $a$. Then we get "another" continuous functional calculus, a star isomorphism $\Gamma_B:C(\sigma_B(a))\cong C^*(a,1_B)$ with $\Gamma_B(1)=1_B$ and $\Gamma_B(z)=a$.
By spectral invariance, we have that $\sigma_A(a)\cup\{0\}=\sigma_B(a)\cup\{0\}$. By the above, we have that if $p$ is a polynomial in $z,\bar{z}$ with constant term $0$, then $\Gamma_A(p)=\Gamma_B(p)$. Now if $f$ is a continuous function defined on $\sigma_A(a)\cup\{0\}$ satisfying $f(0)=0$, then by Weierstrass approximation we can find a sequence of polynomials $(p_n)$ in $z,\bar{z}$  such that $p_n\to f$ uniformly on $\sigma_A(a)\cup\{0\}$. Thus $\Gamma_A(f)=\Gamma_B(f)$. In other words, the continuous functional calculus when performed on functions that fix the origin is independent of the $C^*$-subalgebra we are working in.
Is this correct? If not, what is wrong? If it is correct, please look at the following. I feel very awkward about this result:
Suppose that we have $C^*$-algebras $B\subset A$ and suppose that $B$ is unital with unit $1_B$. Now let $b\in B$ be a positive element with $\varepsilon1_B\leq b$, for some $\varepsilon>0$. Then $\sigma_B(b)\subset[\varepsilon,\|b\|]$, so $b$ is invertible in $B$, since $0$ is not in the spectrum. Moreover, $\sigma_A(b)\cup\{0\}=\sigma_B(b)\cup\{0\}$. If $0\not\in\sigma_A(b)$ (which implies that $A$ is unital and that $b$ is invertible in $A$) then $\sigma_A(b)=\sigma_B(b)\subset[\varepsilon,\|b\|]$. We can define $f(t)=1/t$ on $\sigma_A(b)=\sigma_B(b)$ and set $f(0)=0$. This defines a continuous function on $\sigma_A(b)\cup\{0\}=\sigma_B(b)\cup\{0\}$ and by the above, $\Gamma_A(f)=\Gamma_B(f)$, i.e. the inverse of $b$ in $B$ is the same as the inverse of $b$ in $A$. More specifically, $1_B=1_A$, which seems odd to me.
We just proved that: $B\subset A$ is an inclusion of unital $C^*$-algebras (the units being $1_B$ and $1_A$ respectively) and we have an element $b\in B$ with $\varepsilon1_B\leq b$ and also $b$ is invertible in $A$. Then $1_B=1_A$.
IF my understanding of the continuous functional calculus is correct, can you give me a more compelling argument to convince me that the reasoning in the above proof is not flawed?
 A: Your arguments are right, but there is indeed a "catch". And the "catch" is the assumption "If $0\not\in\sigma_A(b)$..." that is in the assumpion that $b$ is invertble in $A$.
This will not happen unless $1_A = 1_B$, in which case it obviously implies that the inverses are the same.
To see that, consider $C = \operatorname{span} (B \cup \{1_A\})$ which is a $C^*$ subalgebra of $A$ with unit $1_A$. If $b$ had an inverse in $A$, then this inverse would be an element of $C$ (further explanation below). So this inverse would be of the form $x + r1_A$ for some $x \in B$ and $r \in \mathbb{C}$. But then
$$1_A = b (x+r1_A) = bx +rb \in B,$$
which implies that the unit $1_A$ of $A$ is an element of $B$. But the $C^*$ albegra $B$ must have a unique unit, hence, $1_A = 1_B$.
The fact that the inverse of $b$ in $A$ is an element of $C$ follows again from the functional calculus, as then $0 \notin \sigma_A(b)$ and $f(t) = 1/t$ is a limit of polynomials with zero constant term.
This closely relates to the following observation: if $B$ is a unital $C^*$ algebra and $A$ is its unitization, then no element of $B$ is invertible in $A$.
