I want to show that if $F$ is a flat $R$-module, then for any $R$-homomorphism $\varphi: M \rightarrow N$, we have $$\ker (1_F \otimes \varphi) \cong F \otimes \ker \varphi $$
The $\supset$ direction is trivial. For the $\subset$ direction, I use the fact that $F\otimes(-)$ preserves the injectivity of the map $$ M /\ker \varphi \overset{\tilde{\varphi}}{\rightarrow} N $$ Thus if $f\otimes m \in \ker (1_F \otimes \varphi)$, i.e. $f\otimes [m]$ mapped to $0$ under $1_F \otimes \tilde{\varphi}$, we have $f\otimes [m] =0$. This suggests me to complete the proof by showing $$ F\otimes \frac{M}{\ker\varphi} \cong \frac{F\otimes M}{F\otimes \ker\varphi} $$ Is this relation true? It seems very plausible for me that $f\otimes[m] \mapsto [f\otimes m]$ gives the isomorphism. But I struggle at the injectivity part of the proof.
Since $F$ is flat, tensoring with it preserves exactness of all exact sequences; from $$ 0\to\ker f\to M\xrightarrow{f} N \to \operatorname{coker}f\to 0 $$ you get the exact sequence $$ 0\to F\otimes\ker f\to F\otimes M\xrightarrow{1_F\otimes f} F\otimes N \to F\otimes\operatorname{coker}f\to 0 $$ which is (isomorphic to) the standard kernel-cokernel sequence for $1_F\otimes f$. Hence $$ \ker(1_F\otimes f)\cong F\otimes\ker f $$