Identification of Gelfand spectrum, $\Sigma(\mathcal{A})$, with $\sigma(A)$, spectrum of Banach Algebra element

Question

I am reading Mathematical Structure of Quantum Mechanics A short Course for Mathematicians by F. Strocchi.

My question is about the text after Proposition 1.5.3. The proposition is that

For a commutative Banach algebra $\mathcal{A}$ there is a one to one correspondence between the set $\Sigma(\mathcal{A})$ of multiplicative linear functionals and proper maximal ideals of $\mathcal{A}$. Furthermore, given $A\in \mathcal{A}$, $\lambda \in \sigma(A)$ iff there exists a multiplicative linear functional $m \in \Sigma(\mathcal{A})$ such that $m(A) = \lambda$.

Afterwards Strocchi calls $\Sigma(\mathcal{A})$ the Gelfand spectrum of $\mathcal{A}$ and claims

Indeed, if $\mathcal{A}$ is generated by a single element $A$, i.e. the linear span of the powers of $A$ is dense in $\mathcal{A}$, then $\Sigma(\mathcal{A})= \sigma(A)$.

I don't understand this very last statement that $\Sigma(\mathcal{A})=\sigma(A)$. I see that there might be a correspondence between elements of $\Sigma(\mathcal{A})$ and elements of $\sigma(A)$ but I can't see how these two sets could be equal. My issue is that elements of $\sigma(A)$ are complex numbers, $\lambda$, (such that $A-\lambda \mathbb{1}$ is not invertible) whereas elements of $\Sigma(\mathcal{A})$ are multiplicative linear functionals on $\mathcal{A}$, i.e. not necessarily complex numbers.

Is the statement that $\Sigma(\mathcal{A})=\sigma(A)$ just making the point that there is a one to one correspondence between the two sets so we might as well call them equal. Perhaps since there is a homomorphism involved it is justified/conventional? I work in physics so I don't know what might be conventional in mathematics texts. My stronger suspicion is that I'm missing something.

Answer 1

You're not really missing anything here. The statement $\Sigma(\mathcal{A})=\sigma(A)$ is just a sloppy way of saying there is a canonical bijection between these two sets (which in fact is a homeomorphism for their natural topologies). It is not exactly normal for mathematicians to write something like this (we would more properly write $\Sigma(\mathcal{A})\cong \sigma(A)$ instead), but this kind of sloppiness is reasonably common.

Explicitly, the canonical bijection is defined as follows: given $m\in\Sigma(\mathcal{A})$, we map it to $m(A)$, which is an element of $\sigma(A)$.