Prove the Ring Homomorphism is Surjective Prove the following homomorphism is surjective.
$$f : Z → \frac{Z}{3Z} × \frac{Z}{5Z}$$ 
I completely get the questions and i can prove it by working out a corresponding pre-image for all of $\frac{Z}{3Z} × \frac{Z}{5Z}$.
But how do i give a formal proof if i change the question and ask 
for any $(p,q)=1$ Prove the following homomorphism is surjective.
$$f : Z → \frac{Z}{pZ} × \frac{Z}{qZ}$$ 
I feel we may have to use Bezout's identity but don't know how to. 
any help appreciated 
Thanks
 A: Hint: Use the Chinese remainder theorem to show that $f: {Z/pZ} \times {Z/qZ} \to Z/(pqZ)$ is bijective. That's all you should need.
A: Yes, one can immediately read off a CRT solution directly from the Bezout identity:
$\quad jp + kq = 1\,\Rightarrow\,\begin{align}{kq\equiv 1\pmod p}\\jp\equiv 1\pmod q\end{align}\,\ $ thus $\,\ x = bjp + akq \ \Rightarrow \begin{align}{x\equiv a\pmod p}\\x\equiv b\pmod q\end{align}$ 
i.e. $\ {\rm mod}\ (p,q)\!:\,\ \begin{align}kq \equiv (1,0)\\ jp\equiv (0,1)\end{align}\,\Rightarrow\, (a,b) = a(1,0)+b(0,1) \equiv akq + bjp $
which reveals the innate linearity, i.e. how the general solution $\,(a,b)\,$ can be generated as a linear combination of the "basis" solutions $\,(1,0)\,$ and $\,(0,1).\,$  This will become clearer when one studies the ring-theoretic form of CRT (and modules).
A: Since $\ker f = 15 \mathbb Z$ has index $15$, the image of $f$ has size $15$ and so $f$ is surjective.
The same argument works in general when $\gcd(p,q)=1$.
(Invoking the Chinese remainder theorem sounds wrong to me because the argument above gives the simplest proof of the Chinese remainder theorem when the ring is $\mathbb Z$.)
