Let $R$ be a ring, $M$ a left $R$-module and $I$ a two-sided ideal in $R$. $M/IM$ is a left $R$-module via
$$ \begin{align} r\cdot (m+IM)=rm+IM \quad \text{ for } r\in R, m\in M. \end{align} $$
Assume $M/IM$ is generated by $\{\bar{m_1}, ..., \bar{m_t}\}$ as a left $R$-module. Let $m_i\in M$ such that $\bar{m_i}=m_i+IM$ in $M/IM$.
Show that $\{m_1,...,m_t\}$ generates $M$ as a left $R$-module.
I already found a proof for this here but it uses Nakayama’s Lemma. I'm trying to prove this without this lemma.
My ideas so far:
Let $$\begin{align}\varphi: M\rightarrow M/IM, \quad m\mapsto m+IM. \end{align}$$
Take $m\in M$. Since $M/IM$ is a finitely generated left $R$-module, there exist $r_1,...,r_t \in R$ such that
$$\begin{align} \varphi(m)&=r_1 \bar{m_1}+...+r_t \bar{m_t}\\ &=r_1(m_1+IM)+...+r_t(m_t+IM)\\ &=r_1m_1+...+r_tm_t+IM. \end{align}$$
Then $m-(r_1m_1+...+r_tm_t)\in IM$ and thus, it exists an $a\in I, n\in M$ such that
$$\begin{align} m=r_1m_1+...+r_tm_t+an. \end{align}$$
And that's the point where I don't know what to do next.
What could I do next? Am I even on the right track?