Tensoring the exact sequence by a faithfully flat module I have problems to do the exercise 1.2.13 from Liu's "Algebraic Geometry and Arithmetic curves".
First, Liu defines that if $M$ is a flat module over a ring $A$, then $M$ is faithfully flat over $A$ if it satisfies one of the following three equivalent conditions:
Let $M$ be a flat module. Then $M$ is $\textit{faithfully flat}$ if 


*

*$M\ne \mathfrak mM$ for every maximal ideal $\mathfrak m$ of $A$.

*Let $N$ be an $A$-module. If $M\otimes_AN=0$, then $N=0$.

*Let $f:N_1\to N_2$ be a homomorphism of $A$-modules. If $f_M:N_1\otimes M\to N_2\otimes M$ is an isomorphism, then so is $f$.


In the text Liu proves the equivalence of these three conditions. 
Now, the exercise 1.2.13 asks the following:
Let $M$ be a faithfully flat $A$-module. Let $N'\to N \to N''$ be a sequence of $A$-modules. Why do it is exact if and only if $N'\otimes M\to N\otimes M\to N''\otimes M$ is exact?

Although this is the usual property that defines the faithfully flatness, (one of the) Liu's definition(s) which is helpful here is the following: 

Let $M$ be a flat $A$-module. Then $M$ is faithfully flat iff for any $A$-module $N$ such that $M\otimes_AN=0$ we have $N=0$.

Recall that $N'\stackrel{u}\to N\stackrel{v}\to N''$ is exact iff $\operatorname{im}u=\ker v$. 
Suppose that $N'\otimes M\stackrel{u\otimes 1_M}\to N\otimes M\stackrel{v\otimes 1_M}\to N''\otimes M$ is exact. Then $(v\otimes 1_M)\circ(u\otimes 1_M)=0$, that is, $(v\circ u)\otimes 1_M=0$. This implies that $v(u(N'))\otimes M=0$, so $v(u(N'))=0$, that is, $v\circ u=0$. Thus $\operatorname{im}u\subseteq\ker v$.
But we know more, namely $\ker(v\otimes 1_M)=\operatorname{im}(u\otimes 1_M)$. Let's use it. If $n\in\ker v$, then $n\otimes x\in\ker(v\otimes 1_M)$ for all $x\in M$, so $(\ker v)\otimes M\subseteq\ker(v\otimes 1_M)=\operatorname{im}(u\otimes 1_M)=(\operatorname{im}u)\otimes M$. It follows $(\ker v/\operatorname{im}u)\otimes M=0$, so $\ker v=\operatorname{im}u$.
