Gelfand transform is a contractive unital homomorphism

For $\chi \in \mathrm{\Delta }(A)$ we have $\chi (1)=1$ and $\mathrm{\mid }\chi (a)\mathrm{\mid }\le \mathrm{\parallel }a\mathrm{\parallel }$ for every $a$, by [L1].

Each $\widehat{a}$ is a continuous complex-valued function on $\mathrm{\Delta }(A)$, since $\widehat{a}$ is the evaluation map $e_a$ and the evaluation topology makes all $e_a$ continuous; thus $\mathrm{\Gamma }(a)\in C(\mathrm{\Delta }(A))$ and it makes sense to speak of $\mathrm{\parallel }\mathrm{\Gamma }(a){\mathrm{\parallel }}_{\mathrm{\infty }}$.

$\mathrm{\Gamma }$ is complex-linear and multiplicative: for all $\chi \in \mathrm{\Delta }(A)$, $\mathrm{\Gamma }(\lambda a+\mu b)(\chi )=\chi (\lambda a+\mu b)=\lambda \chi (a)+\mu \chi (b)=(\lambda \mathrm{\Gamma }(a)+\mu \mathrm{\Gamma }(b))(\chi )$ and $\mathrm{\Gamma }(ab)(\chi )=\chi (ab)=\chi (a)\chi (b)=(\mathrm{\Gamma }(a)\mathrm{\Gamma }(b))(\chi )$, by linearity and multiplicativity of characters.

$\{\chi (a):\chi \in \mathrm{\Delta }(A)\}={\sigma }_A(a)$ and $\mathrm{\mid }\chi (a)\mathrm{\mid }\le r(a)\le \mathrm{\parallel }a\mathrm{\parallel }$ for every $\chi$, by [L3].

$\mathrm{\parallel }\mathrm{\Gamma }(a){\mathrm{\parallel }}_{\mathrm{\infty }}=r(a)$ for every $a$: the set of values $\{\widehat{a}(\chi ):\chi \}=\{\chi (a):\chi \}$ equals ${\sigma }_A(a)$ by [step 1.4], and by [step 1.1] and [L1] the function $\widehat{a}$ is bounded with $\mathrm{\mid }\widehat{a}(\chi )\mathrm{\mid }\le \mathrm{\parallel }a\mathrm{\parallel }$, so the supremum over $\chi$ of $\mathrm{\mid }\widehat{a}(\chi )\mathrm{\mid }$ is the maximum of $\mathrm{\mid }z\mathrm{\mid }$ over $z\in {\sigma }_A(a)$, that is, $r(a)$; in particular $\mathrm{\parallel }\mathrm{\Gamma }(a){\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }a\mathrm{\parallel }$.

$\mathrm{\Gamma }(1)=\mathbf{1}$, the constant function one: $\mathrm{\Gamma }(1)(\chi )=\chi (1)=1$ for every $\chi$ by [step 1.1].

By [step 1.3], [step 2.1] and [step 2.2], $\mathrm{\Gamma }$ is a unital algebra homomorphism with $\mathrm{\parallel }\mathrm{\Gamma }(a){\mathrm{\parallel }}_{\mathrm{\infty }}=r(a)\le \mathrm{\parallel }a\mathrm{\parallel }$ for all $a$; hence it is contractive.