Verify $f: \mathbb{Z} ⟶ \mathbb{Z}_{r} \times \mathbb{Z}_{s}$ given by $a ⟶ (ā_r, ā_s)$ is an epimorphism

Let $$r,s \in \mathbb{Z}^+$$ such that $$(r,s)=1.$$ Consider the ring $$\mathbb{Z}_{r} \times \mathbb{Z}_{s}$$ (with respect to ordinary multiplication and addition, component by component), prove that the application $$f: \mathbb{Z} ⟶ \mathbb{Z}_{r} \times \mathbb{Z}_{s}$$ given by $$a ⟶ (ā_r, ā_s)$$ is an epimorphism of rings.

I've already proved that it's a homomorphism since

$$(a+b)$$ maps to $$([a+b]_r, [a+b]_s)$$, which is $$([a]_r + [b]_r) + ([a]_s + [b]_s)$$.

$$(ab)$$ maps to $$([ab]_r, [ab]_s)$$, which is $$([a]_r [b]_r) ([a]_s [b]_s$$)$. These proposition are true owning to the operations defined in modular arithmetic. So it's a homomorphism. If the map is surjective then for $$(x,y) \in ℤ_r \times ℤ_s$$ there's an $$a \in ℤ$$. This means that for any $$(x,y)$$, the equations $$[a]_r = x$$, and $$[a]_s = y$$ are simultaneously solvable. But I'm failing at showing the latter proposition. • Hint: Chinese remainder theorem. Jan 30, 2021 at 16:15 • ok, so we have$a ≡ x \quad mod\, r \Longleftrightarrow [a]_r = x$, and$a ≡ y \quad mod\, s \Longleftrightarrow [a]_s = y$, and these are simultaneously solvable by crt since$r,s\$ are coprime? Jan 30, 2021 at 16:24
• Yes, precisely. Jan 30, 2021 at 16:39
• I believe that this is part of the proof of the Chinese Remainder Theorem (for arbitrary rings), hence it feels a little bit cheap to invoke the Chinese Remainder Theorem to prove that this homomorphism is surjective. I have given a more general proof below. Jan 30, 2021 at 17:37
• Remark: A ring epimorphism is not the same as a surjective ring homomorphism. A surjective ring homomorphism is always a ring epimorphism, but the converse is not true. Jan 30, 2021 at 20:38

Given any relatively prime integers $$r$$ and $$s,$$ consider the ring homomorphism $$\pi : \mathbb Z \to (\mathbb Z / r \mathbb Z) \times (\mathbb Z / s \mathbb Z)$$ defined by $$\pi(n) = (n + r \mathbb Z, n + s \mathbb Z).$$ By hypothesis that $$r$$ and $$s$$ are relatively prime, Bézout's Lemma implies that there exist integers $$a$$ and $$b$$ such that $$ra + sb = 1.$$ Consequently, for any element $$(x + r \mathbb Z, y + s \mathbb Z)$$ of $$(\mathbb Z / r \mathbb Z) \times (\mathbb Z / s \mathbb Z),$$ we have that $$(x + r \mathbb Z, y + s \mathbb Z) = (xra + xsb + r \mathbb Z, yra + ysb + s \mathbb Z) = (xsb + r \mathbb Z, yra + s \mathbb Z).$$ Can you find an element $$n$$ of $$\mathbb Z$$ such that $$\pi(n) = (xsb + r \mathbb Z, yra + s \mathbb Z)$$ to finish the proof?
Like I indicated in my comment above, the original question is part of the proof of the Chinese Remainder Theorem for arbitrary rings; the only thing to do is to replace "relatively prime integers" with "comaximal ideals $$I$$ and $$J.$$" From here, it follows that there exist elements $$i \in I$$ and $$j \in J$$ such that $$i + j = 1_R,$$ and the rest follows similarly to the above proof.