I want to show that $\mathbb Z/pq\mathbb Z\cong \mathbb Z/p\mathbb Z\times \mathbb Z/q\mathbb Z$ where $(p,q)=1$. I consider the morphism $$[a]_{pq}\longmapsto ([a]_p,[a]_q).$$
If I have the injectivity, then, I'll have the surjectivity (since the two set has same cardinality).
method 1 : So $$([a]_p,[a]_q)=0\implies p\mid a\wedge q\mid a\underset{(p,q)=1}{\implies} pq\mid a.$$ How can I get $a=0$ ?
method 2
I consider $$([a]_p,[b]_q)\longmapsto [ab]_{pq}.$$ So $$[ab]_{pq}=0\implies pq\mid ab$$ and since $(p,q)=1$, we get $p\mid ab$ and $q\mid ab$. How can I conclude that $a=b=0$ ?
In your first method, if $pq\mid a$, it mean that $a=0$ (in $\mathbb Z/pq\mathbb Z$). So, it's finish.
In your second method, your map is even not a group homomorphism since $$\varphi\Big[([a],[b])+([c],[d])\Big]=[(a+c)(b+d)]\neq [ab]+[cd]=\varphi([a],[b])+\varphi([c],[d]).$$