Let $s$ and $t$ be positive integers greater or equal to $2$ such that $\gcd(s,t)=1$, and consider the function $$\phi:\mathbb{Z}_{st} \rightarrow \mathbb{Z}_s \oplus \mathbb{Z}_t \quad \text{given by} \quad \phi(x)=(x\bmod s, x\bmod t)$$
Prove that $\phi$ is well-defined and an isomorphism whenever $\gcd(s,t)=1$.
$\textbf{First to prove it is well-defined.}$
If $a \bmod (st)=b \bmod (st)$ then we know they share the same remainder so $st \mid a-b$ thus $a-b=st(x)$ for some integer $x$ by the associative property $a-b=s(tx)$ and by the commutative and associative property $a-b=t(sx)$ therefore $s\mid a-b$ and $t \mid a-b$ therefore $a\bmod s=b\bmod s$ and $a\bmod t=b \bmod t$.
The converse holds as long as $s$ and $t$ are relatively prime.
We know $s(x)=t(y)$. Now suppose $\gcd(s,t)=1$. By Bezout $s(n)+t(m)=1$ for some integers $n,m$. Then $t \mid x$ or $x\mid t(m)$ for some integer $m$, $a-b=x(s)=(tm)(s)=st(m)$ thus $st\mid a-b$ therefore $a \mod st=b\mod st$.
$\textbf{Now to show that $\phi$ is injective.}$
Suppose $\phi(x)=\phi(y)$ then $x \equiv y \bmod s$ and $x \equiv y \bmod t$. Thus $x \equiv y \bmod (st)$ since $\gcd(s,t)=1$ by the Chinese remainder theorem.
$\textbf{Now to show that $\phi$ is surjective.}$
Since the the Euler phi function is multiplicative meaning $\phi(st)=\phi(s)\phi(t)$ then the cardinality of $\mathbb{Z}_{st}$ and $\mathbb{Z}_s \oplus \mathbb{Z}_t$ are the same. Thus we have a injective function between two finite sets with the same cardinality. Therefore by the $2$ out of $3$ principle we can conclude that the function is also surjective thus a bijection.
$\textbf{Now to show that $\phi$ is a homomorphism.}$
To show $\phi$ is a homomorphism; because for all $n, m \in \mathbb{Z}$ we have \begin{align*} \phi((n)_{st} + (m)_{st}) &= \phi((n+m)_{st}) = ((n+m)_s, (n+m)_t) = ((n)_s + (m)_s, (n)_t + (m)_t) \\ &= ((n)_s, (n)_t) + ((m)_s, (m)_t)=\phi((n)_{st}) + \phi((m)_{st}). \end{align*} Then $\phi$ is operation preserving.