Let $A$ be a unital commutative Banach Algebra.
Let $\Omega(A)$ be the set of non-zero homomorphisms from $A$ to $\Bbb C$.
Suppose the set of idempotents $N$ in $A$ have dense linear span.
Show the map $f:\Omega(A)\rightarrow \{0,1\}^N$ defined by $f(\phi)(e) = \phi(e)$ is injective.
In The spectrum of a commutative unital Banach algbra having linearly dense idempotents is totally disconnected, it says that the injectivity comes from the dense linear span. But I don't really see how this goes.
So if $f(\phi)(e) =f (\psi)(e)$ , we have $\phi(e)=\psi(e)$. Why does the dense linear span tell us $\phi=\psi$?
Thank you for your help!
If $\phi(e)=\psi (e)$, for every idempotent $e$, then $\phi(a)=\psi (a)$, for every element $a$ which can be written as a linear combination of idempotents (because $\phi$ and $\psi $ are linear).
The set of all such $a$ is suposedly dense, so $\phi$ and $\psi $ coincide on a dense set, and hence everywhere because they are continuous.