Let $U_k(n) = \{x \in U(n) : x \bmod k = 1\}$. For some integers $s, t$ such that $\gcd(s, t) = 1$, show that $\beta : U_s(st) \to U(t)$ defined by $\beta(x) = x \bmod t$ is an isomorphism.
Edit: I have attempted showing injectivity by having some $x, y \in U_s(st)$ such that $\beta(x) = \beta(y) \implies x \bmod t = y \bmod t$. From here, I suppose that $t$ divides $x - y$. I'm thinking that the fact that $x, y \in U_s(st)$ may have something to do with this. Am I correct in saying that $\gcd(x, s) = \gcd(y, s) = 1$? And if so, does that have to do anything with the problem?
For surjectivity, I was assuming that it might be a similar application of the Chinese Remainder Theorem. The day before, I was shown a proof that $\alpha: U(st) \to U(s) \oplus U(t)$ defined by $\alpha(x) = (x \bmod s, x \bmod t)$ was an isomorphism, utilizing the Chinese Remainder Theorem. Is it applicable here?
For showing that it's a homomorphism, I don't believe I need help for that portion of the question.
Edit 2: Here is my proposed proof for the injectivity of $\beta$.
Suppose that $\beta(x) = \beta(y)$ for some $x, y \in U_s(st)$. Then $x \bmod t = y \bmod t \implies x - y \equiv 0 \bmod t \implies t|x - y$. Because any $x \in U_s(st)$ means that $x \bmod s = 1$, then $x \bmod s = y \bmod s \implies s|x - y$. Because $\gcd(s, t) = 1$, then $st|x - y \implies x \equiv y \bmod st$. Because $x, y \in U_s(st)$, then $x = y$. Hence $\beta$ is injective.