Proving that a function is a bijection

Let $m_1,\ldots,m_n$ be pairwise coprime and let $m=m_1m_2\cdots m_n$. Show that the map

\begin{align}
\theta\,\colon \mathbb{Z}_m &\to \prod_{i=1}^n \mathbb{Z}_{m_i}\\
a+m\mathbb{Z} &\mapsto (a_1+m_1\mathbb{Z},\ldots,a_n+m_n\mathbb{Z})
\end{align}

is an isomorphism of rings assuming you know it is well defined and a homomorphism of rings.

---

So it is left to show that the function is a bijection. So if we take

$$(a_1+m_1\mathbb{Z},\ldots,a_n+m_n\mathbb{Z})\in \prod_{i=1}^n \mathbb{Z}_{m_i}$$

Since each $(m_i,m_j)$ are pairwise coprime for $i\not=j$, by the Chinese remainder theorem, there exists a unique $x$ such that

$$x\equiv a_i\pmod{m_i}$$

since this is true for any $1\le i\le n$ we have that

$$x\equiv a\pmod{m}\implies x = a + m\mathbb{Z}$$

So there exists an inverse map:

\begin{align}
\theta^{-1}\,\colon \prod_{i=1}^n \mathbb{Z}_{m_i} &\to \mathbb{Z}_m\\
(a_1+m_1\mathbb{Z},\ldots,a_n+m_n\mathbb{Z}) &\mapsto a+m\mathbb{Z}
\end{align}

Hence $\theta$ is surjective. Since $x$ is unique, it is also injective. Therefore $\theta$ is bijective.

---

Is this a sufficient proof for such a question?

Answer

It is left to show that the function is a bijection. Take

$$(a_1+m_1\mathbb{Z},\ldots,a_n+m_n\mathbb{Z})\in \prod_{i=1}^n \mathbb{Z}_{m_i}$$

Since each $(m_i,m_j)$ are pairwise coprime for $i\not=j$, by the Chinese remainder theorem, there exists a unique $x$ such that

$$x\equiv a_i\pmod{m_i}$$

since this is true for any $1\le i\le n$ we have that

$$x\equiv a\pmod{m}\implies x = a + m\mathbb{Z}$$

So there exists an inverse map:

\begin{align}
\theta^{-1}\,\colon \prod_{i=1}^n \mathbb{Z}_{m_i} &\to \mathbb{Z}_m\\
(a_1+m_1\mathbb{Z},\ldots,a_n+m_n\mathbb{Z}) &\mapsto a+m\mathbb{Z}
\end{align}

Hence $\theta$ is surjective. Since $x$ is unique, it is also injective. Therefore $\theta$ is bijective.