By Lemma \ref{lemma-limit-module-essentially-finite-presentation} and its proof we can find a system $R_\lambda \to S_\lambda$ of local ring maps together with maps of $S_\lambda$-modules $u_\lambda : M_\lambda \to N_\lambda$ satisfying the conclusions (1) -- (6) for both $N$ and $M$ of that lemma and such that the colimit of the maps $u_\lambda$ is $u$. By Lemma \ref{lemma-colimit-eventually-flat} we may assume that $N_\lambda$ is flat over $R_\lambda$ for all sufficiently large $\lambda$. Denote $\mathfrak m_\lambda \subset R_\lambda$ the maximal ideal and $\kappa_\lambda = R_\lambda / \mathfrak m_\lambda$, resp.\ $\kappa = R/\mathfrak m$ the residue fields. \medskip\noindent Consider the map $$ \Psi_\lambda : M_\lambda/\mathfrak m_\lambda M_\lambda \otimes_{\kappa_\lambda} \kappa \longrightarrow M/\mathfrak m M. $$ Since $S_\lambda/\mathfrak m_\lambda S_\lambda$ is essentially of finite type over the field $\kappa_\lambda$ we see that the tensor product $S_\lambda/\mathfrak m_\lambda S_\lambda \otimes_{\kappa_\lambda} \kappa$ is essentially of finite type over $\kappa$. Hence it is a Noetherian ring and we conclude the kernel of $\Psi_\lambda$ is finitely generated. Since $M/\mathfrak m M$ is the colimit of the system $M_\lambda/\mathfrak m_\lambda M_\lambda$ and $\kappa$ is the colimit of the fields $\kappa_\lambda$ there exists a $\lambda' > \lambda$ such that the kernel of $\Psi_\lambda$ is generated by the kernel of $$ \Psi_{\lambda, \lambda'} : M_\lambda/\mathfrak m_\lambda M_\lambda \otimes_{\kappa_\lambda} \kappa_{\lambda'} \longrightarrow M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'}. $$ By construction there exists a multiplicative subset $W \subset S_\lambda \otimes_{R_\lambda} R_{\lambda'}$ such that $S_{\lambda'} = W^{-1}(S_\lambda \otimes_{R_\lambda} R_{\lambda'})$ and $$ W^{-1}(M_\lambda/\mathfrak m_\lambda M_\lambda \otimes_{\kappa_\lambda} \kappa_{\lambda'}) = M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'}. $$ Now suppose that $x$ is an element of the kernel of $$ \Psi_{\lambda'} : M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'} \otimes_{\kappa_{\lambda'}} \kappa \longrightarrow M/\mathfrak m M. $$ Then for some $w \in W$ we have $wx \in M_\lambda/\mathfrak m_\lambda M_\lambda \otimes \kappa$. Hence $wx \in \Ker(\Psi_\lambda)$. Hence $wx$ is a linear combination of elements in the kernel of $\Psi_{\lambda, \lambda'}$. Hence $wx = 0$ in $M_{\lambda'}/\mathfrak m_{\lambda'} M_{\lambda'} \otimes_{\kappa_{\lambda'}} \kappa$, hence $x = 0$ because $w$ is invertible in $S_{\lambda'}$. We conclude that the kernel of $\Psi_{\lambda'}$ is zero for all sufficiently large $\lambda'$! \medskip\noindent By the result of the preceding paragraph we may assume that the kernel of $\Psi_\lambda$ is zero for all $\lambda$ sufficiently large, which implies that the map $M_\lambda/\mathfrak m_\lambda M_\lambda \to M/\mathfrak m M$ is injective. Combined with $\overline{u}$ being injective this formally implies that also $\overline{u_\lambda} : M_\lambda/\mathfrak m_\lambda M_\lambda \to N_\lambda/\mathfrak m_\lambda N_\lambda$ is injective. By Lemma \ref{lemma-mod-injective} we conclude that (for all sufficiently large $\lambda$) the map $u_\lambda$ is injective and that $N_\lambda/u_\lambda(M_\lambda)$ is flat over $R_\lambda$. The lemma follows.