I am reading Wikipedia and there is something I don't understand:
''Let $A $ be a unital commutative Banach algebra over $\mathbb C$. Since $A $ is then a commutative ring with unit, every non-invertible element of $A$ belongs to some maximal ideal of $A$. Since a maximal ideal $\mathfrak m$ in $A$ is closed, $A/ \mathfrak m$ is a Banach algebra that is a field, and it follows from the Gelfand-Mazur theorem that there is a bijection between the set of all maximal ideals of $A$ and the set $Δ(A)$ of all nonzero homomorphisms from $A$ to $\mathbb C$. ''
I understand everything up to and including the Gelfand-Mazur theorem. What I don't get is how every character corresponds to a maximal ideal. I know that because of Gelfand-Mazur that if $\mathfrak m$ is maximal then there is an isomorphism $\varphi : A/\mathfrak m \to \mathbb C$. True then that this $\varphi $ (with slight modification) is a character. But I don't see how this yields a bijection -- a map defined in this way will not be surjective as it does not map to characters what aren't surjective. And surely, there is nothing to prevent a character $\chi$ to have $\mathrm{im}(\chi) \subsetneq \mathbb C$?
You are working with $\mathbb{C}$-Banach algebras and $\mathbb{C}$-linear homomorphisms. In particular, if $\chi:A\to \mathbb{C}$ is a character then $\chi(c\cdot 1_A) = c\cdot \chi(1_A) = c\cdot 1 = c$ for any $c\in \mathbb{C}$, i.e., characters are surjective.