Theorem. If $\fm_1,\ldots,\fm_n$ are ideals in a commutative ring $R$ such that $\fm_i+\fm_j=R$ for all $i\neq j$, then the natural map $f:R\to\prod_i R/\fm_i$ is surjective.
Proof. I'll illustrate for $n=3$---the exact same argument applies for any $n$. By hypothesis there are elements $a_1,a_2,b_1,b_3,c_2,c_3$ such that
$$\begin{matrix} a_1\in\fm_1, & a_2\in\fm_2, & a_1+a_2=1 \\ b_1\in\fm_1, & b_3\in\fm_3, & b_1+b_3=1 \\ c_2\in\fm_2, & c_3\in\fm_3, & c_1+c_3=1 \end{matrix}$$
Thus their images under $f$ are as such:
$$\begin{matrix} f(a_1)=(0,1,*), & f(a_2)=(1,0,*) \\ f(b_1)=(0,*,1), & f(b_3)=(1,*,0) \\ f(c_2)=(*,0,1), & f(c_3)=(*,1,0) \end{matrix}$$
The asterisks denote unknowns. To prove surjectivity it suffices to construct preimages of $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. Well, it can be clearly seen that
$$(1,0,0)=f(a_2b_3),\quad (0,1,0)=f(a_1c_3),\quad (0,0,1)=f(b_1c_2).$$
This completes the proof. $\qed$