I tried to prove:
Let $A$ be a commutative unital complex Banach algebra. Then there is a bijection between the maximal ideals in $A$ and the set of non-zero homomorphisms $A \to \mathbb C$.
But I can't do the last step of the proof. Here is what I have so far:
Let $\chi : A \to \mathbb C$ be a character. Then $\mathrm{ker}(\chi)$ is a maximal ideal in $A$. Then $\mathfrak m$ is closed and hence $A/ \mathfrak m$ is a Banach algebra in which every non-zero element is invertible. Let $\varphi$ denote the map that takes $\chi$ to $\mathrm{ker}(\chi)$. Let $\psi$ be the map taking a maximal ideal $\mathfrak m \subseteq A$ to the homomorphism $A \to \mathbb C$ corresponding to the Gelfand-Mazur isomorphism $A / \mathfrak m \to \mathbb C$.
Then $\varphi$ is a bijection because $\psi$ is a left and a right inverse of $\varphi$:
$$ \varphi (\psi (\mathfrak m)) = \varphi (f_{\mathfrak m}: A \to \mathbb C ) = \mathfrak m$$
is clear hence $\psi $ is a right inverse of $\varphi$.
$$ \psi (\varphi (\chi )) = \psi (\mathrm{ker}(\chi)) = f_{\mathrm{ker}(\chi)}: A \to \mathbb C$$
where $f$ is the map mapping $a$ to the unique element in $\sigma(a)$ or $0$ if $a$ is in the kernel of $\chi$.
How to show that $f_{\mathrm{ker}(\chi)} = \chi$?