linear map between finitely generated free modules (over local ring) which become injective after tensoring with the residue field Let $F,G$ be finitely generated free modules over a Noetherian local ring $(R,\mathfrak m)$. Let $f: F\to G$ be an $R$-linear map such that the induced map $\bar f: F/\mathfrak m F \to G/\mathfrak m G$ is injective i.e. $F\cap f^{-1}(\mathfrak m G)=\mathfrak m F.$
Then, is it true that $f$ is split injective?
My thoughts: Let $L=\ker (f)$. Then clearly $L \subseteq F\cap f^{-1}(\mathfrak m G)=\mathfrak m F$. But to show $L=0$, I need the stronger inclusion $L\subseteq \mathfrak m L$, which I am unable to show. Finally, I have no idea why $f$ would split (why the image of $f$ is a direct summand of $G$?)
 A: Assume WLOG that $F=R^n$ and $G=R^m$, so that $f\colon R^n\rightarrow R^m$ is represented by an $m\times n$-matrix with entries in $R$ and the induced map $\overline{f}\colon(R/\mathfrak{m}R)^n\rightarrow(R/\mathfrak{m}R)^m$ is represented by the $m\times n$-matrix with entries in $R/\mathfrak{m}R$ obtained by elementwise reduction modulo $\mathfrak{m}$ from this matrix. The injectivity of $\overline{f}$ implies that we can find a projection $\overline{p}\colon(R/\mathfrak{m}R)^m\rightarrow(R/\mathfrak{m}R)^n$ onto a subset of factors s.t. $\overline{p}\circ\overline{f}\colon(R/\mathfrak{m}R)^n\rightarrow(R/\mathfrak{m}R)^n$ is an isomorphism. Choosing projection on the same subset of factors, we obtain a projection map $p\colon R^m\rightarrow R^n$ that lifts $\overline{p}$, so that $p\circ f\colon R^n\rightarrow R^n$ lifts $\overline{p}\circ\overline{f}$. The naturality of the determinant now tells us that $\det(p\circ f)+\mathfrak{m}=\det(\overline{p}\circ\overline{f})$. Since $\overline{p}\circ\overline{f}$ is an automorphism of $(R/\mathfrak{m}R)^n$ and $R/\mathfrak{m}R$ is a field, linear algebra tells us that $\det(\overline{p}\circ\overline{f})$ is non-zero, which means that $\det(p\circ f)\not\in\mathfrak{m}$. Since $R$ is local, it follows that $\det(p\circ f)\in R^{\times}$, which implies that $p\circ f$ is invertible. Then, $(p\circ f)^{-1}\circ p\colon R^m\rightarrow R^n$ is a retraction for $f$.
