Characters in Banach algebras

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.
• @Fermat There is nothing special about $1_A$ in this answer. Any non-zero $\mathbb{C}$-linear transformation $\phi:V\to \mathbb{C}$ with $V$ a $\mathbb{C}$-vector space is surjective: if $\phi(v)\neq 0$, then $c = \phi(\phi(v)^{-1} c v)$ for any $c\in \mathbb{C}$. – Jan Ladislav Dussek Feb 2 at 15:06
• Thank you, you are right, and what about injectivity? Specially for characters on Banach algebra $A$. (i.e., multiplicative linear functionals) – Fermat Feb 2 at 17:05