In Functional Analysis, we have covered Functional Calculus, that is, a way to associate, once having fixed a Banach algebra $A$ and an element $a\in A$, an element $\tilde f(a)\in A$ to every $f$ holomorphic in a neighborhood of the spectrum of $a$, via an integral formula, in such a way that $f\mapsto\tilde f(a)$ is an algebra homomorphism, $\tilde f(a)=e$ (the identity) if $f\equiv1$, $\tilde f(a)=a$ if $f(z)=z$, the subalgebra of $\tilde f(a)$'s is commutative even if $A$ is not, and these properties characterize this $f\mapsto \tilde f(a)$ map. On one occasion, the professor, being asked if this mapping was injective, replied:
Uuuh, è iniettiva, certo, fino a quando, ovviamente, tengo fisso l'elemento, quando pói cambio l'elemento, chi lo sa.
i.e.:
Uuuh, is it injective, of course, up till, of course, I keep the element fixed, when I change the element, who knows.
I smell rat here. After all, $\mathbb{C}$ is a B.A., and this mapping is certainly not injective, in general, since $f(z)=z$ is certainly not the only holomorphic map sending $z_0$ e.g. into itself, yet this mapping is clearly $f\mapsto f(z_0)$, in this case, if the fixed $a$ from above is $z_0$. So what did she mean? Are there conditions under which this is injective indeed? Perhaps as soon as the spectrum is not a single point for each element (i.e. as soon as we leave $\mathbb{C}$) this mapping becomes injective?
To determine whether the map is injective, we first need to clarify the domain. If we let the domain be $\mathscr{O}(\sigma(a)) = \bigcup_{\sigma(a) \subset U} \mathscr{O}(U)$, where the $U$ are open sets, then the map $\Phi \colon f \mapsto \tilde{f}(a)$ is trivially not injective, we can let $U = D_1 \cup D_2$ consist of two disjoint open disks, with $\sigma(a) \subset D_1$, and let
$$f(z) = \begin{cases} 0 &, z \in D_1 \\ 1 &, z \in D_2. \end{cases}$$
Then clearly $\tilde{f}(a) = 0$ but $f \neq 0$.
For an interesting question, we need to ignore the behaviour of $f$ "far away" from $\sigma(a)$. Thus we look at $\mathscr{O}(\sigma(a))/{\sim}$, where $f \sim g$ if and only if there is an open $W \supset \sigma(a)$ with $f\lvert_W = g\lvert_W$. Note that we can also describe $\mathscr{O}(\sigma(a))/{\sim}$ as the inductive limit of the Banach algebras $A(U) = \mathscr{C}(\overline{U}) \cap \mathscr{O}(U)$ where $U$ ranges over the bounded neighbourhoods of $\sigma(a)$.
Then $\Phi$ is always injective when $\sigma(a)$ contains no isolated points: By linearity of $\Phi$, we need to prove $\tilde{f}(a) = 0$ for an $f \in \mathscr{O}(U)$ implies $f \sim 0$, i.e. $f$ vanishes on a neighbourhood of $\sigma(a)$. The spectral mapping theorem gives us $\{0\} = \sigma(0) = \sigma(\tilde{f}(a)) = f(\sigma(a))$, so $f\lvert_{\sigma(a)} \equiv 0$ for $\tilde{f}(a) = 0$. Let $W$ be a connected component of $U$ intersecting $\sigma(a)$, and $w \in W\cap \sigma(a)$. Since $\sigma(a)$ contains no isolated points, it follows that $w$ is an accumulation point of $W \cap \sigma(a)$, and by the identity theorem, $f\lvert_W \equiv 0$. Thus if $V$ is the union of the components of $U$ intersecting $\sigma(a)$, we have $f\lvert_V \equiv 0$, and since clearly $\sigma(a) \subset V$, indeed $f \sim 0$.
If $\sigma(a)$ contains isolated points, then generally $\Phi$ is not injective, for $f\lvert_{\sigma(a)} \equiv 0$ does then not imply $f \sim 0$.
If the closed subalgebra of $A$ generated by $a$ satisfies $\sigma(x) = \{0\} \implies x = 0$, then $\Phi$ is never injective if $\sigma(a)$ has isolated points, for then $\tilde{f}(a) = 0 \iff f\lvert_{\sigma(a)} \equiv 0$, and if $p$ is an isolated point of $\sigma(a)$, we can take
$$f(z) = \begin{cases} (z - p) &, \lvert z-p\rvert < \delta \\ \quad 0 &, \lvert z-p\rvert > \delta,\end{cases}$$
where $0 < \delta < \operatorname{dist}(p,\sigma(a)\setminus \{p\})$ as an $f$ with $\tilde{f}(a) = 0$ but $f \nsim 0$.
Question: Can we have $f \mapsto \tilde{f}(a)$ injective even if $\sigma(a)$ has isolated points? Answer: Yes.
Let $A = \mathscr{B}(\ell^2(\mathbb{N}))$, and consider
$$a \colon (x_0, x_1, x_2,\dotsc) \mapsto \bigl( 0, \tfrac{1}{1}x_0, \tfrac{1}{2} x_1, \tfrac{1}{3} x_2, \dotsc).$$
Then $\lVert a^k\rVert = \frac{1}{k!}$, so $\sigma(a) = \{0\}$. To see that $\Phi$ is injective for that $a$, consider a convergent power series
$$f(z) = \sum_{n = 0}^{\infty} c_n z^n.$$
Then we have
$$\tilde{f}(a) = \sum_{n = 0}^{\infty} c_n a^n,$$
since the series on the right converges absolutely in $\mathscr{B}(\ell^2(\mathbb{N}))$. If $f \neq 0$, let $m = \min \{ n \in \mathbb{N} : c_n \neq 0\}$. Then
$$\langle \tilde{f}(a) e_0, e_m\rangle = \frac{c_m}{m!} \neq 0$$
and consequently $\tilde{f}(a) \neq 0$, showing that $\Phi$ is injective.