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?
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}.