prop 2.19 (ii)=>(i) of Atiyah Macdonald I know that there are some post about this question.
But since none of them gave me answer, I again post it here.
What I want to show is "every short exact sequence tensored by N is again a short exact sequence" implies "N is a flat module".
(cf) Atiyah defines "N is falt" <=> "If A->B->C is exact at B, then A⨂N->B⨂N ->C⨂N is exact at B⨂N")
My approach is 
But since the canonical map from Im(g)⨂N->Im(g⨂I) is not in general injective, my approach fails. Can somebody help me?
 A: Now I got to know how to handle it.
Assume ii) holds. To prove i) amounts to show that
$$A\overset{f}{\to}B\overset{g}{\to}C
\label{ses}\tag{1}$$
exact implies
$$A\otimes N
\overset{f\otimes 1}{\to}B\otimes N
\overset{g\otimes 1}{\to}C\otimes N$$
exact, i.e., that $\operatorname{Im}(f\otimes 1)=\operatorname{Ker}(g\otimes 1)$, where $1$ is the identity on $N$.
Suppose then that \eqref{ses} is exact. We have the following short exact sequences:
$$
\begin{align}
0&\to\operatorname{Im} f
\overset{i}{\to}B
\overset{g'}{\to}\operatorname{Im}g
\to 0,\\
0&\to\operatorname{Im} g
\overset{j}{\to}C
\to\operatorname{Coker}g
\to 0,
\end{align}
$$
and by ii), it follows that the sequences
$$
\begin{align}0&\to\operatorname{Im} f \otimes N
\overset{i\otimes 1}{\to}B \otimes N
\overset{g'\otimes 1}{\to}\operatorname{Im}g \otimes N
\to 0,\\
0&\to\operatorname{Im} g \otimes N
\overset{j\otimes 1}{\to}C \otimes N
\to\operatorname{Coker}g \otimes N
\to 0 \tag{2} \label{second_ses}
\end{align}
$$
are exact too.
Hence, $\operatorname{Ker}(g'\otimes 1)=\operatorname{Im}(i\otimes 1)=\operatorname{Im}(f\otimes 1)$ (check the last identity). Therefore, if we show that
$$\operatorname{Ker}(g'\otimes 1)=\operatorname{Ker}(g\otimes 1),
\tag{3}\label{last_eq}$$
the proof will end. Why is \eqref{last_eq} then true? Well, exactness of \eqref{second_ses} implies in particular that $j\otimes 1$ is injective, and since in general the identity
$(\alpha'\circ\alpha) \otimes (\beta'\circ\beta)
=(\alpha'\otimes\beta') \circ (\alpha\otimes \beta)$ holds (see the last two paragraphs of Sect. Tensor product of modules from Atiyah-MacDonald's book), we have that
$$
(j\otimes 1)\circ (g'\otimes 1)
= (j\circ g')\otimes 1
=g\otimes 1
$$
and so
$$
\operatorname{Ker}(g\otimes 1)
=\operatorname{Ker}((j\otimes 1)\circ (g'\otimes 1))
=\operatorname{Ker}(g'\otimes 1),
$$
where in the last equality we have used the injectivity of $j\otimes 1$. That is, we have \eqref{last_eq}.
