Proof
(related to Lemma: Coprimality and Congruence Classes)
 By hypothesis, $m$ is a positive integer and $a(m)$ is a given congruence class.
 Assume, we have two representatives $a_1,a_2$, i.e. $a_1(m)\equiv a_2(m).$
 By definition, this means $m\mid (a_1a_2).$
 For any divisor $d\mid m$ we have by the divisibility law no. 3 that $d\mid (a_1a_2).$
 By the divisibility law no. 9, this is equivalent to $d\mid a_1$ and $d\mid a_2.$
 Therefore, $d$ is a common divisor of $m,$ $a_1$ and $a_2.$
 Since the greatest common divisor is unique^{1}, it follows $\gcd(a_1,m)=\gcd(a_2,m).$
 In particular, this is true if $a_1\perp m$ and $a_1\perp m$ are coprime, i.e. $\gcd(a_1,m)=\gcd(a_2,m)=1.$
 Therefore, any representative $a\in\mathbb Z$ of a given congruence class $a(m)\in\mathbb Z_m$ is either coprime or not coprime to $m.$
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References
Bibliography
 Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927
Footnotes