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I was going to prove:

Let $A$ be a unital Banach algebra. Then $$\sigma(a) = \{\tau(a) \mid \tau \in \Omega (A)\}$$

and I started the following argument:

Let $\lambda \in \sigma (a)$ and let $\tau$ be any character. Then $\tau ( a - \lambda) = 0$ since homomorphisms map invertible elements to invertible elements. Then $\tau (a) = \tau (\lambda) = \lambda$.

I don't see the mistake but there obviously is one since this "argument" shows that all characters are equal to all $\lambda $ at $a$.

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    $\begingroup$ Homomorphisms map invertible elements to invertible elements, but they can also map some non-invertible elements to invertible elements. Consider an evaluation homomorphism on $C([0,1])$, just because $f$ has a zero, it doesn't follow that $f\equiv 0$. $\endgroup$ – Daniel Fischer Nov 6 '14 at 22:51
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It is true that homomorphisms map invertible elements to invertible elements, but in general, they also map some non-invertible elements to invertible elements.

If we look at $C([0,1])$ and $\tau \colon f \mapsto f(1)$ for example, we have $\tau(\operatorname{id}) = 1$, but $\operatorname{id}$ is not invertible since it has a zero.

So the mistake is that you conclude $\tau(a-\lambda) = 0$ from $\lambda\in \sigma(a)$.

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