Flat and faithful (as functor) implies faithfully flat I'm trying to understand the proof of $(2) \Rightarrow (1)$
in this lemma in The Stack Project.
But I couldn't follow the argument. How could I see $\alpha \otimes \mathrm{id}_M$ is zero by the exactness of the complex $-\otimes_R M$? Can you give me some hints? Thanks

 A: We have $$\alpha\otimes id_M: R\otimes_R M\to N_2/Im(N_1)\otimes_R M$$ sending $$\alpha\otimes id_M(r\otimes  m)=\alpha(r)\otimes m=(rx+Im(N_1))\otimes m.$$
$rx\otimes m\in N_2\otimes_R M$ such that under the map $N_2\otimes_R M\to N_3\otimes_R M$ it maps $rx\otimes m\mapsto 0$.
So there is $\beta=\sum n_i\otimes m_i \in N_1\otimes M$ such that $\beta\mapsto rx\otimes m$ under the map $f_1\otimes id_M: N_1\otimes_R M\to N_2\otimes_R M$.
So
$rx\otimes m=(f_1\otimes id_M)(\beta)=\sum f_1(n_i)\otimes m_i\in N_2\otimes M$ so that under the map $$N_2\otimes M\to N_2/Im(N_1)\otimes_R M$$
$$rx\otimes m\mapsto 0$$
Hence $\alpha\otimes id_M(r\otimes  m)=0$.
A: Let me write $f:N_1\to N_2$ and $g: N_2\to N_3$.
The functor $-\otimes_R M$ preserves cokernels and there is a map
$$\ker(g)\otimes_R M \to \ker(g\otimes 1),$$
and then $\alpha\otimes 1$ is obtained by looking at the composition
$$\ker(g)\otimes_R M \to \ker(g\otimes 1) \subseteq N_2\otimes_R M \to \operatorname{coker}(f\otimes 1) = \operatorname{coker}(f)\otimes_R M.$$
Now $\ker(g\otimes 1) = \operatorname{im}(f\otimes 1)$ since the tensored sequence is exact, which means that each of its elements has zero class in $\operatorname{coker}(f\otimes 1)$. This shows that $\alpha\otimes 1=0$.
A: Hint:
$rx\in \ker(N_2\longrightarrow N_3)=\operatorname{Im}N_1 $.
