 Reduction map on unit groups

Reduction modulo an ideal $I$ of a commutative ring $R$ gives a group homomorphism $R^{\times }\to (R/I{)}^{\times }$. It need not be surjective: the units $\{1,-1\}$ of $\mathbb{Z}$ do not lift the classes two and three in $(\mathbb{Z}/5\mathbb{Z}{)}^{\times }$.

A useful sufficient condition is $I\subseteq J(R)$, the Jacobson radical. If $ab\equiv 1(\mathrm{mod}I)$, then $ab$ is a unit by the unit criterion for the Jacobson radical, and consequently $a$ is a unit. Every invertible residue class then lifts.

For positive integers $n\mid m$, reduction $(\mathbb{Z}/m\mathbb{Z}{)}^{\times }\to (\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ is always surjective. Start with a representative coprime to $n$ and use the Chinese remainder theorem to choose a congruent representative avoiding each additional prime dividing $m$. This explicit integer result should be distinguished from unrestricted reduction in an arbitrary ring.