Doubt in understanding the proof that a character of a unital abelian Banach algebra is of norm 1. In the following theorem

Theorem : Let $A$ be a unital abelian Banach algebra.
If $\tau \in \Omega(A)$, then $\|\tau\|=1$.

Proof. If $\tau \in \Omega(A)$ and $a \in A$, then $\bf \tau(a) \in \sigma(a)$, so $|\tau(a)| \leq r(a) \leq\|a\|$. Hence, $\|\tau\| \leq 1 .$ Also, $\tau(1)=1$, since $\tau(1)=\tau(1)^{2}$ and $\tau(1) \neq 0 .$ Hence, $\|\tau\|=1$
I cannot understand how $$\tau(a) \in \sigma(a)$$
Note : A character on an abelian algebra $A$ is a non-zero homomorphism $\tau: A \rightarrow \mathbf{C}$. It is denoted by $\Omega(A)$ the set of characters on $A$. And $\sigma(a)$ is the spectrum of $a$.

If $a-\tau(a)$ were invertible with inverse $b$, i.e., $(a-\tau(a)1)b=1$, then
$$
0=(\tau(a)-\tau(a))\tau(b)=\tau((a-\tau(a)1)b)=\tau(1)=1,
$$
which is of course impossible.
