Let $A$ be a commutative Banach algebra and let $$G: A \rightarrow C_{0}(\text{Spec}(A))$$ be a topologically injective map.
Recall that the map $T: X \rightarrow Y$ is called topologically injective if there exists a positive $C > 0$ such that $||Tx|| \geq C ||x||$ for all $x \in X$. One can also show that the latter is equivalent to the definition in which $$T: X \rightarrow \text{im}(T)$$ is a homeomorphism (in fact, it is the original definition)
How to prove that the Gelfand map $G$ is topologicall injective if and only if there exists $C > 0$ such that for all $a \in A$ $$||a^{2} || \geq C ||a||^{2}$$
The spectral radius $r(a)=\lim_{n\to\infty} \|a^{2^n}\|^{\frac{1}{2^n}}\geq \lim_{n\to\infty} (C^n\|a\|^{2^n})^{\frac{1}{2^n}}=\|a\|$, thus $\|G(a)\|=r(a)=\|a\|$. So, $G$ is in fact an isometry.