Suppose $R$ is a commutative ring and $I$ is a maximal ideal. Let $M = R^m$ and $N=R^n$ be finitely generated $R$-modules. Suppose $\varphi: M \to N$ and $\phi: M/IM \to N/IN$ are module homomorphisms, where $\phi(\overline{m}) = \overline{\varphi(m)}$.
Prove that if $\varphi$ (and thus $\phi$) is surjective, then $n \leq m$.
Why do we need to use $\phi$? Isn't it impossible for every $(b_1,\dots,b_n) \in R^n$ to have a corresponding $(a_1,\dots,a_m) \in R^m$ such that $\varphi\bigg((a_1,\dots,a_m)\bigg) = (b_1,\dots,b_n)$ if $n \not\leq m$.
The modules $\bar{M}=M/IM$ and $\bar{N}=N/IN$ are in a natural fashion vector spaces over the field $R/I$ and $\phi$ is a linear map.
Note that $\bar{M}$ is generated by $m$ elements, namely $\overline{e_i}=e_i+IM$, $i=1,2,\dots,m$, where $e_i$ is the element of $R^n$ having $1$ in the $i$-th component and zero elsewhere.
Suppose $\overline{a_1}=a_1+I,\dots,\overline{a_m}=a_m+I\in R/I$ and $$ \overline{0}=\sum_{i=1}^m \overline{a_i}\,\overline{e_i} $$ This translates into $$ \sum_{i=1}^m a_ie_i=(a_1,a_2,\dots,a_m)\in IR^n $$ Thus $a_i\in I$ and therefore $\overline{a_i}=\overline{0}$, for $i=1,2,\dots,m$.
Hence $\{\overline{e_1},\overline{e_2},\dots,\overline{e_m}\}$ is a basis for $\bar{M}$ and $\dim\bar{M}=m$. Similarly, $\dim\bar{N}=n$. The result then follows from the standard rank-nullity theorem for vector spaces.