I need to explicitly calculate the inverse of
$$
\mathbb{Z}/89081\mathbb{Z}\rightarrow\mathbb{Z}/229\mathbb{Z} \times \mathbb{Z}/389 \mathbb{Z}
$$
with
$$
a \mod 89081 \mapsto (a\mod 229, a\mod 389)
$$

Now I knew this: $\mathbb{Z}/229\mathbb{Z}$ and $\mathbb{Z}/389$ are cyclic and if an isomorphism exits between $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/mZ \rightarrow \mathbb{Z}/nm\mathbb{Z}$ with the groups cyclic, this is equivalent to $\gcd(n,m)=1$, so I know that $\gcd(229,389)=1$. Now I suspect that it has something to do with finding the inverse of the element modulo, I am not sure, maybe you could give me a hint. It needs to be such an $a$ that the inner product is $a^{-1}$. What do I need to do to get there? Do I have to calculate the inverse of $229,389 \mod 89081$? And where to next?